Lesson 27: Projectiles Launched at an Angle

To do questions involving objects launched from the ground upwards at an angle (like kicking a football up into the air and watching it as it arcs in the air and comes back down), you need to add a few more steps to the way you did the questions for objects launched horizontally in Lesson 26.

Vertical Velocity Method

Imagine for a moment that you are watching an object as it rises into the air after you kick it upwards at an angle. Look at Figure 1 below as you read through this description.

Figure 1

We can use this information about its vertical movement to make some calculations. We know...

This gives us enough information to calculate the maximum height of the flight, and the time it spends in the air. After that, we can calculate just about anything…

Example 1: You kick a soccer ball at an angle of 40° above the ground with a velocity of 20m/s.

  1. How high will it go?
  2. How much time does it spend in the air?
  3. How far away from you will it hit the ground (aka range)?
  4. What is the ball’s velocity when it hits the ground?

Before we can calculate anything else, we first need to break the original velocity into components.

We do this so we have a vertical component to do the first couple calculations with. The horizontal component will be used later.

Figure 2

sinΘ = opp/hyp
opp = sinΘ (hyp)
= sin 40° (20m/s)
opp = vy = 13m/s


At its maximum height, halways through its flight, the object won't be going up or down, so we'll say that its final velocity at that point is zero.

vf2 = vi2 + 2ad
d = (vf2 - vi2) / 2a
= (02 - 132) / 2(-9.81)
d = 8.6 m

The ball will reach a maximum height of 8.6 m.


Same ideas as above...

a = (vf - vi) / t
t = (vf - vi) / a
= (0 - 13) / -9.81
t = 1.3 s

But this is only the time to the halfway point, so the final answer is 2.7s.


It is moving at a constant velocity horizontally during the whole time we just figured out, so let’s use Figure 2 that was drawn above to get the horizontal component of the velocity.

cosΘ = adj/hyp
adj = cos Θ (hyp)
= cos 40° (20m/s)
opp = vy = 15m/s

v = d/t
d = vt
= 15m/s (2.7s)
d = 41 m

41m is the horizontal distance (range) that it has traveled.

d) The ball’s velocity when it hits the ground is exactly the same as when it was originally launched… 20 m/s at 40° up from the horizontal. The only difference is that now it's spiking into the ground.

Vertical Displacement Method

To do the problem this way, we must assume that the object strikes the ground at exactly the same height that it was launched from. This is the case for most questions.

When you look at how the object has moved in Figure 1, you’ll notice that it has a total vertical displacement of zero.

Example 2: Redo part (b) of the example that we did above. The other parts should still be done the same way (basically).


Use the vertical velocity we calculated above as necessary. The vertical displacement is zero (from start to end), so…

d = vit + 1/2 at2
0 = vit + 1/2 at2
-vit = 1/2 at2 <- divide both sides by "t"
-vi = 1/2 at <- solve for "t"
t = -2vi / a
= -2(13) / -9.81
t = 2.6 s

The answer for time using this method doesn’t have to be doubled, since it is already for the entire flight.