I like to call the next type of projectile motion angled projectile motion. In this
type, an object is projected upward at an angle and then returns back to the same height
it was at originally. The neat thing about angled projectile motion is that it lands at
the same angle at which it came off the surface.
As before in horizontal projectile motion, the horizontal
component of the velocity remains constant while the vertical undergoes a constant
acceleration of -9.8m/s2. And like in vertical motion, since the beginning and ending
points of travel are the same vertically, the velocities will be the same in magnitude
but only opposite in direction. For example, if you kick a football giving it a velocity
of 15m/s at 60° then it will land with a velocity of 15m/s at -60°. Continuing with the
football situation, let's do a sample problem.
A football is kicked with a speed of 15m/s at an angle of 30°.
a. How far does it travel until it hits the ground?
b. How long is it in flight?
c. How high does it rise?
First we need to find the horizontal and vertical components. If you are unclear as
to how to do this please refer back to vector lesson 2.
15m/s cos 60° = 7.5m/s = x component
15m/s sin 60° = 12.99m/s = y component
As before, the first thing we need to is find the time. For this we can use the
acceleration formula.
-9.8m/s2 = (-12.99m/s - 12.99m/s) / t
-9.8m/s2 = -25.98m/s / t
2.651s = t
The fact that because we know the y component, we know the initial and final vertical
velocities and can use them in a formula like this makes that property very useful. Now
that we know part b, let's go back and get part a.
7.5m/s = d / 2.651s
19.883m = d
And finally, on to part c. Since we know that the football will be at it's highest
point halfway through the trip, we must cut the time in half and then use
d= vot + 0.5 * a * t2 to find its greatest height.
0.5 * 2.651s = 1.326s
d = 12.99m/s * 1.326s + 0.5 * -9.8m/s2 * (1.326s)2
d = 8.609m
As is with most introductory physics problems, the math is relatively easy. The
tough part is figuring out what to do. As a change from what we've been doing, instead
of starting off with the speed and angle, let's start off with the height and speed and
let you find the angle.
A long jumper jumps at a 20° angle and attains a maximum altitude of 0.6m.
a. What is her initial speed?
b. How far is her jump?
Since we have to work backwards in this problem, let's plug what we already know
into the equations.
We know that v sin 20° = vo,so we just need to plug in
v sin 20° for vo in the equation 2ad = vf2 - vo2.
2 * -9.8m/s2 * 0.6m = (0m/s)2 - (v sin 20°)2
-11.76m2/s2 = -(v sin 20°)2
11.76m2/s2 = (v sin 20°)2
sqrt (11.76m2/s2) = v sin 20°
3.429m/s = v sin 20°
10.027m/s = v
Now we find the horizontal and vertical components to help us find part b.
10.027m/s cos 20° = 9.422m/s
10.027m/s sin 20° = 3.429m/s
Then we find the time, and multiply that times the velocity to the the total distance
of the trip.
-9.8m/s2 = (-3.429m/s - 3.429m/s) / t
.6998s = t
.6998s * 9.422m/s = d
6.594m = d
See, even working backwards is easy if you know where to start.
Lesson 3 will cover bi-level projectile motion.
