We've already learned about the equations for average
velocity and average acceleration.
Those equations are useful, but they don't give exact values for velocity or acceleration, just an
average.
In the position vs. time graph below, each path from A to B has a different acceleration, but they all
have some properties in common.
Example: How much time does each path from A to B take?
solution
They all take the same time.
$$\Delta t = t_f - t_i$$
$$\Delta t = 50\,\mathrm{s}-5\,\mathrm{s}$$
$$\Delta t = 45\,\mathrm{s}$$
Example: What is the displacement for each path.
solution
Each path travels a different distance, but they all have the same displacement.
$$\Delta x = x_f - x_i$$
$$\Delta x = 25\,\mathrm{m}-5\,\mathrm{m}$$
$$\Delta x = 20\,\mathrm{m}$$
Example: What is the average velocity for each path?
solution
Each path ends up with the same total displacement over the same period of time.
This means they all have the same average velocity.
The average velocity equals the displacement divided by the time period.
$$v_{\mathrm{avg}} = \frac{\Delta x}{\Delta t}$$
$$v_{\mathrm{avg}} = \frac{25\, \mathrm{m}-5\, \mathrm{m}}{50\, \mathrm{s}-5\, \mathrm{s}}$$
$$v_{\mathrm{avg}} = \frac{20\, \mathrm{m}}{45\, \mathrm{s}}$$
$$v_{\mathrm{avg}} = 0.\overline{44} \, \mathrm{\frac{m}{s}}$$
Question: What is different about each path?
answer
Each path has a different:
acceleration
initial velocity
final velocity
distance traveled
color
acceleration =
m/s²
There are many ways to move from A to B, but if acceleration is limited to a constant value only one path works.
When graphed as position vs time the path is a parabola.
Question: Set the acceleration to 0.05 m/s².
Write a short description of the object's velocity as it moves from A to B.
answer
(at the default A and B positions)
The velocity is always increasing by 0.05 m/s every second.
The initial velocity directed away from point B.
Starting at point A, the object slows down to a stop.
The object then speeds up towards point B until it arrives.
Constant Acceleration
When acceleration is constant we can predict position and velocity at any point in time.
These predictions come from the equations of motion.
derivation of the equations of motion
The first equation is the average acceleration equation rearranged with acceleration as a constant
value.
$$a_{\mathrm{avg}} = \frac{\Delta v}{\Delta t}$$
$$a = \frac{v - u}{\Delta t}$$
$$a \Delta t = v - u$$
$$\large \boxed{v = u + a \Delta t}$$
When acceleration is constant, velocity changes at a constant rate.
This means that the average velocity equals half of the sum of initial and final velocities.
The next one also starts with the average velocity for constant motion equation form above.
We can plug our previous equation into this to remove the final velocity.
$$v = u + a \Delta t$$
$$\Delta t = \frac{v-u}{a}$$
$$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^2 $$
$$\Delta x = u \left(\frac{v-u}{a}\right) + \tfrac{1}{2}a \left(\frac{v-u}{a}\right)^2 $$
$$a\Delta x = u(v-u)+ \tfrac{1}{2}(v-u)^2 $$
$$2a\Delta x = 2u(v-u)+ (v-u)^2 $$
$$2a\Delta x = (2uv-2u^2)+ (v^2 - 2uv + u^2) $$
$$2a\Delta x = v^2 - u^2 $$
$$u^2 = v^2 -2a\Delta x$$
$$ \boxed{v^2 = u^2 +2a\Delta x}$$
$$v = u+a \Delta t$$
$$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$
$$\Delta x = \tfrac{1}{2}(v+u)\Delta t$$
$$v^{2} = u^{2}+2a \Delta x$$
\(\Delta x\) = displacement [m] vector
\(\Delta t\) = time period [s]
\(v\) = final velocity [m/s] vector
\(u\) = initial velocity [m/s] vector
\(a\) = acceleration [m/s²] (constant) vector
Each equation is missing one of the five variables.
Looking for the missing variable can help you choose the right equation for each situation.
Working with multiple equations can be complicated. It helps to follow steps:
List the known and unknown variables.
Choose an equation with only one unknown variable.
Use algebra to isolate the unknown variable on one side of the equation.
Plug the knowns into the equation and simplify.
Check if your answer agrees with your intuition and expected units.
Example: A car moving at 30 m/s puts on its brakes and comes to a stop in 10 meters.
Find the acceleration of the car. Assume the friction from the brakes produces constant acceleration.
solution
A car coming to a quick stop should have a large negative acceleration.
The units are correct for acceleration.
Our answer looks good!
Example: A bullet aimed straight up leaves the barrel of a gun at 400 m/s. It accelerates
down at 9.8 m/s². If the bullet travels for 40.8 s before stopping how far up did it go? Ignore air
friction.
solution
$$u = 400 \, \mathrm{\tfrac{m}{s}}$$
$$a = -9.8 \, \mathrm{\tfrac{m}{s^{2}}}$$
$$\Delta t = 40.8 \, \mathrm{s}$$ $$\Delta x =\, ?$$
$$\Delta x = u\Delta t + \tfrac{1}{2} a \Delta t^{2}$$
$$\Delta x = (400\, \mathrm{\tfrac{m}{s}}) (40.8\, \mathrm{s}) + \tfrac{1}{2} (-9.8 \,
\mathrm{\tfrac{m}{s^{2}}})(40.8 \, \mathrm{s})^{2}$$
$$\Delta x = 16320\, \mathrm{m} - 8157\, \mathrm{m}$$
$$\Delta x = 8163 \, \mathrm{m}$$
Question: What are some situations when acceleration is constant?
answer
free fall with no air friction (a = 9.8 m/s²)
just being at rest (v = 0 m/s) (a = 0 m/s²)
moving at a constant speed (a = 0 m/s²)
Question: What are some situations when acceleration is NOT constant?
answer
speeding up (accelerating) (a > 0)
hitting the ground (a = -9.8 → a > 0 → a = 0)
dancing (a = ?)
Example: A plane takes off at a speed of 170 miles/hour while accelerating from rest on a
runway that is 6000 ft long.
Find the acceleration of the plane in m/s².
On Earth's surface everything is pulled down at
9.8 m/s².
The rate is the same for cars, birds, puppies, apples, balloons, and everything.
g = acceleration from gravity on the Earth's surface = 9.8 m/s²
Effects like air friction, thrust, and buoyancy can change the perceived acceleration of gravity.
To keep things simple I will ignore these effect for the example problems.
Example: A sleeping cat falls from rest off of a ledge. If the cat hits the ground moving
at 6.0 m/s how long was the cat in free fall?
solution
The acceleration of gravity comes from massive objects.
Every planet, star, moon, and asteroid has a different surface gravity.
name
g (m/s²)
Sun
275
Mercury
3.7
Venus
8.9
Earth
9.8
Moon
1.6
Mars
3.7
Ceres
0.27
Jupiter
25.8
Saturn
10.4
Uranus
8.7
Neptune
11.2
Example: Imagine a meteor 4000 m above the Moon falls from rest. What is the impact
velocity of the meteor? How long does it take for the meteor to hit the surface?
solution
Example: You drop a rock from 2.0 meters above the ground. It hits the ground after 1.03
seconds. What planet, moon, or asteroid are you on?
solution
We can solve for the acceleration and compare it to the surface gravities on the chart.
$$a = ?$$
$$\Delta x = -2\,\mathrm{m}$$
$$u = \mathrm{rest} = 0$$
$$\Delta t = 1.03\,\mathrm{s}$$
$$\Delta x = u\Delta t + \tfrac{1}{2} a \Delta t^{2}$$
$$-2 = (0)(1.03) + \tfrac{1}{2}a(1.03)^{2}$$
$$-4 = a(1.03)^{2}$$
$$-4 = a(1.08)$$
$$-3.70 \, \mathrm{\tfrac{m}{s^{2}}}= a$$
It's Mars! (Or Mercury)
2-D Motion (for constant acceleration)
Solving for 2-Dimensional motion can reuse the methods from 1-Dimension if we divide the problem up into 2
directions.
We also should set the two directions to be perpendicular so that they are independent from each other.
This gives each direction unrelated positions, velocities, and accelerations, but they still share the same
time period.
Δt =
Δx =
Δy =
u =
u =
v =
v =
a =
a =
You can organize the variables in columns for x and y with shared time.
Example: An object is moving at 3 m/s north, and it is accelerating at 1 m/s² north.
It is also moving at 5 m/s east, and accelerating at 2 m/s² west.
How far does the object move in 10 seconds?
setup
Let's make the x-direction east-west, and the y-direction north-south, like a compass.
East and north will be positive, while south and west will be negative.
Δt = 10 s
Δx = ?
Δy = ?
u = 5 m/s
u = 3 m/s
v
v
a = -2 m/s²
a = 1 m/s²
solution
We'll start with the north-south, y direction.
$$\Delta y = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$
$$\Delta y = (3)(10) + \tfrac{1}{2}(1)(10)^{2}$$
$$\Delta y = 30 + 50$$
$$\Delta y = 80 \, \mathrm{m}$$
The east-west, x direction can use the same equation.
$$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$
$$\Delta x = (5)(10) + \tfrac{1}{2}(-2)(10)^{2}$$
$$\Delta x = 50 + -100$$
$$\Delta x = -50 \, \mathrm{m}$$
The object moved 80 m north and 50 m west.
We can also calculate the displacement with the Pythagorean theorem.
When solving for objects in free fall on the
surface of the Earth you can let the x-direction be horizontal and the y-direction be vertical.
This choice puts the acceleration from gravity in only the y-direction.
Δt =
↔ horizontal
↕ vertical
Δx =
Δy =
u =
u =
v =
v =
a = 0
a = -9.8 m/s²
Vectors directed down or left are negative and vectors directed up or right are positive.
Example: A ball moving horizontally at 2 m/s rolls off a table that is 1.5 m high.
Find how far the ball travels horizontally before it hits the ground.
setup
It's not possible to solve for the horizontal distance without knowing the time.
If you aren't sure why try plugging the information into an equation of motion.
We can find the time with the vertical information and then use it for the horizontal.
Δt =
Δx = ?
Δy = -1.5 m
u = 2 m/s
u = 0
v =
v =
a = 0
a = -9.8 m/s²
solution
$$\Delta y = u\Delta t + \tfrac{1}{2}a \Delta t^2$$
$$-1.5 = 0\Delta t + \tfrac{1}{2}(-9.8)\Delta t^2$$
$$-1.5 = -4.9\Delta t^2$$
$$-4.9\Delta t^2 = -1.5$$
$$\sqrt{\Delta t^2} = \sqrt{0.31}$$
$$\Delta t = \pm 0.55$$
The negative time answer is valid, but not what we are looking for.
$$\Delta t = 0.55 \, \mathrm{s}$$
With this new information we can use time to solve on the horizontal side.
Δt = 0.55 s
Δx = ?
Δy = 1.5 m
u = 2 m/s
u = 0
v = 2 m/s
v = -5.42 m/s
a = 0
a = 9.8 m/s²
$$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = 2(0.55) + \tfrac{1}{2}(0)(0.55)^{2}$$
$$\Delta x = 1.1 \, \mathrm{m}$$
Aim and click to fire. Please be careful not to hit each other.
Δt =
Δx =
Δy =
u =
u =
v =
v =
a = 0
a = -9.8 m/s²
Question: Which variables stay constant as time changes?
answer
The accelerations and initial velocities stay constant.
Also the final horizontal velocity is constant.
Question: Why do all the shots that land on the ground end with a negative Δy?
answer
$$\Delta y = y_f - y_i$$
Δy is the vertical displacement. It measures the difference between the starting height and the
ending height.
It doesn't matter how high the object goes, if it starts on the ground and ends on the ground the
vertical displacement is going to be zero.
The Δy is negative because the tank's turret is above ground.
When the projectile ends up on the ground, it is lower then the starting point on the turret.
Question: Where on the projectile's arc is its vertical velocity zero?
answer
The vertical velocity is zero at the top of the arc, the highest point.
Example: A projectile has a velocity of 150 m/s.
It is fired at an angle of 30° above the horizon.
Use trigonometry to find the parts of the velocity in the
horizontal and vertical directions.
solution
Example: A ball is thrown at an angle of 60° above the horizon and a speed of 10 m/s.
If the ball is thrown from 2.0 meters above the ground how far does the ball travel in the horizontal
direction before it hits the ground?
strategy
Warning: This is a long complicated example problem. Take your time. Get organized.
Some equations will be dead ends. Other equations will lead to the quadratic equation.
You can avoid using the quadratic equation if you solve for the final vertical velocity before you
solve for time.
Start with the 2 column structure. You already know the accelerations.
Δt =
Δx =
Δy =
u =
u =
v =
v =
a = 0
a = -9.8 m/s²
You can find the x and y part of the initial velocity with trigonometry.
You also know Δy. It's just the difference between the starting and ending point in the vertical
direction.
The path of the ball doesn't matter, just look at the difference.
Example: An arrow at ground level is fired at 60° above the horizon at 10 m/s.
Calculate the maximum height of the arrow.
strategy
At first it might seem like there isn't enough information to find the vertical distance.
The trick is to end the problem at the highest point on the arc which makes the final velocity 0.
In this simulation we can fire boxes at a wall.
Use 2-D kinematics calculations to predict what height the gap in the wall should be to let a box
through.
mouse position = (0,0)
wall horizontal distance = m
initial horizontal velocity = m/s
initial vertical velocity = m/s
gap vertical midpoint = m
Use 2-D kinematics equations to calculate the gap height that will allow a box to pass through.
Measure the wall's distance and height by moving your mouse over the
simulation.
Input your calculation for the vertical midpoint of the gap, and then click fire.
solution
We need to send a box through the gap in the wall again.
This time we can change the initial vertical velocity to get it through.
mouse
position = (0,0)
wall horizontal distance = m
gap vertical midpoint = m
initial horizontal velocity = m/s
initial vertical velocity = m/s
Use 2-D kinematics equations to calculate the initial vertical velocity.
Measure the wall's distance and height by moving your mouse over the
simulation.
Input your calculation for the missing initial vertical velocity, and then click fire.
solution
In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!
Unless a problem says otherwise, assume Earth gravity and zero air friction.
Example: A dynamics cart rolls along a straight track in a classroom. It moves 0.120 km to the right in 15 s while a motion sensor samples its position 50 times each second. What is its average velocity?
solution
$$0.120\,\mathrm{km}=120\,\mathrm{m}$$
$$v_{\mathrm{avg}} = \frac{\Delta x}{\Delta t}$$
$$v_{\mathrm{avg}} = \frac{120\,\mathrm{m}}{15\,\mathrm{s}}$$
$$v_{\mathrm{avg}} = 8.0\,\mathrm{\tfrac{m}{s}}$$
The positive answer means the average velocity is to the right.
Example: A runner leaves the starting line and speeds up from 3 m/s to 21 m/s in 6 s. The track is 400 m around, but only this short interval matters. What is the runner's average acceleration?
solution
$$a_{\mathrm{avg}} = \frac{\Delta v}{\Delta t}$$
$$a_{\mathrm{avg}} = \frac{21\,\mathrm{\tfrac{m}{s}} - 3\,\mathrm{\tfrac{m}{s}}}{6\,\mathrm{s}}$$
$$a_{\mathrm{avg}} = 3.0\,\mathrm{\tfrac{m}{s^2}}$$
Example: A skateboard starts at 14.4 km/h and accelerates at 2.5 m/s² for 8 s. What is its final velocity?
solution
$$14.4\,\mathrm{km/h}=4.0\,\mathrm{\tfrac{m}{s}}$$
$$v = u + a\Delta t$$
$$v = 4.0\,\mathrm{\tfrac{m}{s}} + (2.5\,\mathrm{\tfrac{m}{s^2}})(8\,\mathrm{s})$$
$$v = 24\,\mathrm{\tfrac{m}{s}}$$
Example: A bike is moving at 30 m/s and brakes with an acceleration of -4 m/s² for 5 s. What is its final velocity?
solution
$$v = u + a\Delta t$$
$$v = 30\,\mathrm{\tfrac{m}{s}} + (-4\,\mathrm{\tfrac{m}{s^2}})(5\,\mathrm{s})$$
$$v = 10\,\mathrm{\tfrac{m}{s}}$$
The bike is still moving forward, but more slowly.
Example: A sled starts at 21.6 km/h on a packed snow path and accelerates at 1.5 m/s² for 10 s. The rider has a 4 kg backpack, but the motion information is enough. How far does it move?
solution
$$21.6\,\mathrm{km/h}=6.0\,\mathrm{\tfrac{m}{s}}$$
$$\Delta x = u\Delta t + \tfrac{1}{2}a\Delta t^2$$
$$\Delta x = (6)(10) + \tfrac{1}{2}(1.5)(10)^2$$
$$\Delta x = 60 + 75$$
$$\Delta x = 135\,\mathrm{m}$$
Example: A toy car goes from 5 m/s to 29 m/s in 8 s with constant acceleration. What is its acceleration?
solution
$$v = u + a\Delta t$$
$$a\Delta t = v - u$$
$$a = \frac{v-u}{\Delta t}$$
$$a = \frac{29 - 5}{8}$$
$$a = 3.0\,\mathrm{\tfrac{m}{s^2}}$$
Example: A cart starts at 2 m/s and reaches 20 m/s while accelerating at 3 m/s². How long does this take?
solution
$$v = u + a\Delta t$$
$$a\Delta t = v-u$$
$$\Delta t = \frac{v-u}{a}$$
$$\Delta t = \frac{20-2}{3}$$
$$\Delta t = 6.0\,\mathrm{s}$$
Example: A cart is already moving at 12 m/s. It accelerates at 2 m/s² over a 50 m track. What is its speed at the end of the track?
solution
Time is not given, so use the equation without time.
$$v^2 = u^2 + 2a\Delta x$$
$$v^2 = (12)^2 + 2(2)(50)$$
$$v^2 = 344$$
$$v = 18.5\,\mathrm{\tfrac{m}{s}}$$
Example: A car moving at 100.8 km/h brakes at -7 m/s² until it stops. How much distance does it need to stop?
solution
The final velocity is zero because the car stops.
$$100.8\,\mathrm{km/h}=28\,\mathrm{\tfrac{m}{s}}$$
$$v^2 = u^2 + 2a\Delta x$$
$$(0)^2 = (28)^2 + 2(-7)\Delta x$$
$$0 = 784 - 14\Delta x$$
$$14\Delta x = 784$$
$$\Delta x = 56\,\mathrm{m}$$
Example: A ball is thrown straight upward and is in the air for 4.0 s. Its mass is 0.20 kg. How high does it go above the launch point?
answer
This cannot be solved from the information given as written. The mass is not useful for this kinematics problem, and the total time in the air only gives the height if the ball lands at the same height it was launched from.
Example: A cart moves 90 m in 6 s while accelerating uniformly. Its final velocity is 18 m/s. What was its initial velocity?
solution
Use the equation with average velocity because displacement, time, and final velocity are known.
$$\Delta x = \tfrac{1}{2}(v+u)\Delta t$$
$$90 = \tfrac{1}{2}(18+u)(6)$$
$$90 = 3(18+u)$$
$$30 = 18+u$$
$$u = 12\,\mathrm{\tfrac{m}{s}}$$
Example: A ball is dropped from rest. How far does it fall in 3 s?
solution
Let upward be positive. The ball starts from rest, so u = 0.
$$\Delta y = u\Delta t + \tfrac{1}{2}a\Delta t^2$$
$$\Delta y = (0)(3) + \tfrac{1}{2}(-9.8)(3)^2$$
$$\Delta y = -44.1\,\mathrm{m}$$
The ball falls 44.1 m downward.
Example: A ball is thrown upward at 18 m/s. What is its velocity after 2 s?
solution
$$v = u + a\Delta t$$
$$v = 18 + (-9.8)(2)$$
$$v = -1.6\,\mathrm{\tfrac{m}{s}}$$
The negative sign means the ball is moving downward after 2 s.
Example: A ball is thrown straight upward at 16 m/s. How high above the release point does it rise?
solution
At the highest point, the vertical velocity is zero.
$$v^2 = u^2 + 2a\Delta y$$
$$(0)^2 = (16)^2 + 2(-9.8)\Delta y$$
$$0 = 256 - 19.6\Delta y$$
$$\Delta y = 13.1\,\mathrm{m}$$
Example: A tennis ball is hit straight upward at 24 m/s. How long does it take to reach its highest point?
solution
At the top, the vertical velocity is zero.
$$v = u + a\Delta t$$
$$0 = 24 + (-9.8)\Delta t$$
$$9.8\Delta t = 24$$
$$\Delta t = 2.45\,\mathrm{s}$$
Example: A stone is dropped from a bridge 20 m above the water. What is its velocity just before it hits the water?
solution
Let upward be positive. The displacement is -20 m, and the stone starts from rest.
$$v^2 = u^2 + 2a\Delta y$$
$$v^2 = (0)^2 + 2(-9.8)(-20)$$
$$v^2 = 392$$
$$v = -19.8\,\mathrm{\tfrac{m}{s}}$$
The negative sign means the stone is moving downward.
Example: A rock is dropped from rest from a height of 45 m. How long does it take to hit the ground?
solution
Let upward be positive, so the displacement is -45 m.
$$\Delta y = u\Delta t + \tfrac{1}{2}a\Delta t^2$$
$$-45 = (0)\Delta t + \tfrac{1}{2}(-9.8)\Delta t^2$$
$$-45 = -4.9\Delta t^2$$
$$\Delta t^2 = 9.18$$
$$\Delta t = 3.03\,\mathrm{s}$$
Example: A stone is thrown downward at 5 m/s from a 30 m cliff. What are its impact velocity and fall time?
solution
Let upward be positive. The starting velocity and displacement are both negative.
$$u = -5\,\mathrm{\tfrac{m}{s}}$$
$$a = -9.8\,\mathrm{\tfrac{m}{s^2}}$$
$$\Delta y = -30\,\mathrm{m}$$
$$v^2 = u^2 + 2a\Delta y$$
$$v^2 = (-5)^2 + 2(-9.8)(-30)$$
$$v^2 = 613$$
$$v = -24.8\,\mathrm{\tfrac{m}{s}}$$
Now use the velocity equation.
$$-24.8 = -5 + (-9.8)\Delta t$$
$$\Delta t = 2.02\,\mathrm{s}$$
Example: A ball rolls horizontally off an 1800 cm tall ledge at 12 m/s. How long is it in the air, and how far from the base of the ledge does it land?
solution
Split the motion. The vertical motion gives the time, then the horizontal motion gives the range.
$$1800\,\mathrm{cm}=18\,\mathrm{m}$$
horizontal
vertical
ux = 12 m/s
uy = 0 m/s
ax = 0
ay = -9.8 m/s²
Δx = ?
Δy = -18 m
$$\Delta y = u_y\Delta t + \tfrac{1}{2}a_y\Delta t^2$$
$$-18 = (0)\Delta t + \tfrac{1}{2}(-9.8)\Delta t^2$$
$$\Delta t = 1.92\,\mathrm{s}$$
$$\Delta x = u_x\Delta t$$
$$\Delta x = (12)(1.92)$$
$$\Delta x = 23.0\,\mathrm{m}$$
Example: A marble leaves a tabletop horizontally at 15 m/s. The tabletop is 20 m above the floor. How far from the table does the marble land?
solution
The marble has no initial vertical velocity because it leaves horizontally.
$$\Delta y = u_y\Delta t + \tfrac{1}{2}a_y\Delta t^2$$
$$-20 = (0)\Delta t + \tfrac{1}{2}(-9.8)\Delta t^2$$
$$-20 = -4.9\Delta t^2$$
$$\Delta t = 2.02\,\mathrm{s}$$
Now use the same time horizontally.
$$\Delta x = u_x\Delta t$$
$$\Delta x = (15)(2.02)$$
$$\Delta x = 30.3\,\mathrm{m}$$
Example: A ball is launched at 25 m/s at an angle of 37° above horizontal. What are its initial horizontal and vertical velocity components?
solution
Horizontal uses cosine, and vertical uses sine.
$$u_x = u\cos\theta$$
$$u_x = (25)\cos(37^\circ)$$
$$u_x = 20.0\,\mathrm{\tfrac{m}{s}}$$
$$u_y = u\sin\theta$$
$$u_y = (25)\sin(37^\circ)$$
$$u_y = 15.0\,\mathrm{\tfrac{m}{s}}$$
Example: A projectile starts with horizontal velocity 20 m/s and vertical velocity 15 m/s. Where is it after 2 s?
solution
Use the same time for both directions.
$$\Delta x = u_x\Delta t$$
$$\Delta x = (20)(2)$$
$$\Delta x = 40\,\mathrm{m}$$
$$\Delta y = u_y\Delta t + \tfrac{1}{2}a_y\Delta t^2$$
$$\Delta y = (15)(2) + \tfrac{1}{2}(-9.8)(2)^2$$
$$\Delta y = 10.4\,\mathrm{m}$$
After 2 s, the projectile is 40 m forward and 10.4 m above its launch height.
Example: A ball is launched from the ground at 72 km/h and 30° above horizontal. How long is it in the air, and what is its range?
solution
First convert the launch speed, then split it into components.
$$72\,\mathrm{km/h}=20\,\mathrm{\tfrac{m}{s}}$$
$$u_x = u\cos\theta$$
$$u_x = (20)\cos(30^\circ)$$
$$u_x = 17.3\,\mathrm{\tfrac{m}{s}}$$
$$u_y = u\sin\theta$$
$$u_y = (20)\sin(30^\circ)$$
$$u_y = 10.0\,\mathrm{\tfrac{m}{s}}$$
Because it lands at the same height it was launched from, Δy = 0.
$$\Delta y = u_y\Delta t + \tfrac{1}{2}a_y\Delta t^2$$
$$0 = (10.0)\Delta t + \tfrac{1}{2}(-9.8)\Delta t^2$$
$$0 = \Delta t(10.0 - 4.9\Delta t)$$
$$\Delta t = 2.04\,\mathrm{s}$$
Now use horizontal motion.
$$\Delta x = u_x\Delta t$$
$$\Delta x = (17.3)(2.04)$$
$$\Delta x = 35.3\,\mathrm{m}$$
Example: The same ball is launched from the ground at 20 m/s and 30° above horizontal. What is its maximum height above the launch point?
solution
Only vertical motion matters for maximum height. From the previous example:
$$u_y = 10.0\,\mathrm{\tfrac{m}{s}}$$
At the highest point, vertical velocity is zero.
$$v_y^2 = u_y^2 + 2a_y\Delta y$$
$$(0)^2 = (10.0)^2 + 2(-9.8)\Delta y$$
$$0 = 100 - 19.6\Delta y$$
$$\Delta y = 5.10\,\mathrm{m}$$
Example: A soccer ball is kicked from 150 cm above the ground at 43.2 km/h and 40° above horizontal. How long is it in the air, and how far forward does it travel?
solution
Start by converting the given values and splitting the launch velocity.
$$150\,\mathrm{cm}=1.5\,\mathrm{m}$$
$$43.2\,\mathrm{km/h}=12\,\mathrm{\tfrac{m}{s}}$$
$$u_x = (12)\cos(40^\circ)$$
$$u_x = 9.19\,\mathrm{\tfrac{m}{s}}$$
$$u_y = (12)\sin(40^\circ)$$
$$u_y = 7.71\,\mathrm{\tfrac{m}{s}}$$
The ball lands 1.5 m below where it was kicked, so Δy = -1.5 m.
Use vertical motion to find the final vertical velocity first.
$$v_y^2 = u_y^2 + 2a_y\Delta y$$
$$v_y^2 = (7.71)^2 + 2(-9.8)(-1.5)$$
$$v_y^2 = 88.9$$
$$v_y = -9.43\,\mathrm{\tfrac{m}{s}}$$
The final vertical velocity is negative because the ball is moving downward when it lands.
$$v_y = u_y + a_y\Delta t$$
$$-9.43 = 7.71 + (-9.8)\Delta t$$
$$\Delta t = 1.75\,\mathrm{s}$$
Now use horizontal motion.
$$\Delta x = u_x\Delta t$$
$$\Delta x = (9.19)(1.75)$$
$$\Delta x = 16.1\,\mathrm{m}$$
Example: A water balloon is thrown from a balcony 2.0 m above the ground at 18 m/s and 25° above horizontal. How far from the balcony does it land?
solution
Split the velocity first.
$$u_x = (18)\cos(25^\circ)$$
$$u_x = 16.3\,\mathrm{\tfrac{m}{s}}$$
$$u_y = (18)\sin(25^\circ)$$
$$u_y = 7.61\,\mathrm{\tfrac{m}{s}}$$
The balloon lands 2.0 m below launch height, so Δy = -2.0 m.
Find the final vertical velocity.
$$v_y^2 = u_y^2 + 2a_y\Delta y$$
$$v_y^2 = (7.61)^2 + 2(-9.8)(-2.0)$$
$$v_y^2 = 97.1$$
$$v_y = -9.85\,\mathrm{\tfrac{m}{s}}$$
Then find the time.
$$v_y = u_y + a_y\Delta t$$
$$-9.85 = 7.61 + (-9.8)\Delta t$$
$$\Delta t = 1.78\,\mathrm{s}$$
Now use horizontal motion.
$$\Delta x = u_x\Delta t$$
$$\Delta x = (16.3)(1.78)$$
$$\Delta x = 29.1\,\mathrm{m}$$
Example: A ball has horizontal velocity 14 m/s and initial vertical velocity 9 m/s. A wall is 21 m away. How high above launch height is the ball when it reaches the wall?
solution
The wall distance is horizontal, so use it to find the time.
$$\Delta x = u_x\Delta t$$
$$21 = (14)\Delta t$$
$$\Delta t = 1.50\,\mathrm{s}$$
Now use that same time in the vertical motion.
$$\Delta y = u_y\Delta t + \tfrac{1}{2}a_y\Delta t^2$$
$$\Delta y = (9)(1.50) + \tfrac{1}{2}(-9.8)(1.50)^2$$
$$\Delta y = 13.5 - 11.0$$
$$\Delta y = 2.48\,\mathrm{m}$$
Example: A launcher gives a ball horizontal velocity 12 m/s. A target opening is 18 m away and 4.0 m above launch height. What initial vertical velocity is needed to pass through the opening?
solution
Use horizontal motion first because it gives the time to reach the opening.
$$\Delta x = u_x\Delta t$$
$$18 = (12)\Delta t$$
$$\Delta t = 1.50\,\mathrm{s}$$
Now solve the vertical displacement equation for the needed initial vertical velocity.
$$\Delta y = u_y\Delta t + \tfrac{1}{2}a_y\Delta t^2$$
$$4.0 = u_y(1.50) + \tfrac{1}{2}(-9.8)(1.50)^2$$
$$4.0 = 1.50u_y - 11.0$$
$$15.0 = 1.50u_y$$
$$u_y = 10.0\,\mathrm{\tfrac{m}{s}}$$
Example: A ball is launched with horizontal velocity 16 m/s and vertical velocity 12 m/s. A 6.0 m tall wall is 24 m away. Does the ball clear the wall?
solution
Find the time to reach the wall using horizontal motion.
$$\Delta x = u_x\Delta t$$
$$24 = (16)\Delta t$$
$$\Delta t = 1.50\,\mathrm{s}$$
Find the height at that time.
$$\Delta y = u_y\Delta t + \tfrac{1}{2}a_y\Delta t^2$$
$$\Delta y = (12)(1.50) + \tfrac{1}{2}(-9.8)(1.50)^2$$
$$\Delta y = 18.0 - 11.0$$
$$\Delta y = 6.98\,\mathrm{m}$$
The ball clears the 6.0 m wall by about 0.98 m.
Example: A small launcher fires a ball from level ground with horizontal velocity 18 m/s and vertical velocity 24 m/s. The launch stand is 0.80 m wide, but take the launch and landing heights as the same. How far away does it land?
solution
Since it lands at the same height it starts, use vertical motion to find the nonzero time in the air.
$$\Delta y = u_y\Delta t + \tfrac{1}{2}a_y\Delta t^2$$
$$0 = (24)\Delta t + \tfrac{1}{2}(-9.8)\Delta t^2$$
$$0 = \Delta t(24 - 4.9\Delta t)$$
$$\Delta t = 4.90\,\mathrm{s}$$
Now use horizontal motion.
$$\Delta x = u_x\Delta t$$
$$\Delta x = (18)(4.90)$$
$$\Delta x = 88.2\,\mathrm{m}$$