Projectile Motion

Projectile motion is the way to find out the values of measurements when something (like a marble, ball, car, trampoline, unicycle, banana, etc.) goes off a cliff, or gets thrown into the air. For this section we will assume there is no air resistance, and that the acceleration due to gravity is 10ms^-2On earth it is close to 9·8ms^-2..

Projectile Motion is measured in two parts, horizontal and vertical. Don’t get them mixed up when you are solving a problem.

Projectile motion uses the same formulas in the Speed with Acceleration section. If you don't remember them, here they are again:

Depending on what the question is asking, we would select the correct formula.

There's no use in just sitting around, lets take some miscellaneous objects to the nearest cliff.

But before we throw things off it, let's first consider this: If you have a bowling ball and a banana, and then we drop"drop" refers to the fact that when objects leave our hands, they had an initial velocity of 0ms^-1. If we "throw" something, we would specify how fast it was thrown. them from a large height, what hits the ground first?

The answer: Both!

The reason: We are ignoring air resistance for this section, so objects fall with the acceleration due to gravity. Otherwise, these calculations would be far too hard to include air resistance, and we don't want that do we?

Our first object is a unicycle. It gets dropped from a height of 320m. How long does it take to hit the ground?

This is an straightforward problem because it only has a vertical measurement. So, we need to find the formulaDo you remember the process we used on the Speed with Acceleration page? If you don't, have a look now. that has all the things we need or know.

Initial Speed = 0ms^-1We dropped the unicycle, which means it didn't have any speed before we let go of it.
Time Taken = ?
Distance = 320m
Acceleration = 10ms^-2This is the acceleration due to gravity, mentioned at the start of the page.

With some close observation, you'll figure out that the formula we need is the second one. So all we need to do is put the values in the equation...

320 = 0×t + ·5×10×t²
320 = 0 + 5t²
64 = t²
t = 8s (1 s.f.Don't know what this is? Significant figures are explained in the Before We Begin section.)

So it took 8 seconds for the unicycle to hit the ground below.

Now let's try an example where the object has a horizontal velocity before it tumbles off the cliff.

A clumsy classmate of yours just tripped over their shoelaces (for the 19th time that day) and accidentally threw a stone they were carrying horizontally off the same cliff as Example 1 at 2·5ms^-1. How far away from the cliff will it end up?

This time, the first step we need to take is to calculate how long the stone will take to hit the ground vertically. The answer is the answer to Example 1, because in both examples, the vertical speed of the object was 0ms^-1.

So we already know the stone will take 8 seconds to reach the ground. Now we need to find the horizontal distance. Using the information given, and the initial horizontal speed of the stone, we can write this:

Initial Horizontal Speed = 2·5ms^-1
Time Taken = 8s
Horizontal Distance = ?
Acceleration = 0ms^-2Why is this zero? This is the horizontal acceleration, and the stone isn't accelerating in the horizontal direction.

And we use the second formula again to get this:

d = 2·5×8 + ·5×0×8²
d = 20 + 0
d = 20m (1 s.f.)

Lets try something easier after that.

You take a stone and throw it down a different cliff face at 4·5ms^-1. Its speed before it hits the ground is 29ms^-1. How long did it take to hit the ground?

You should know what to do by now.

Initial Speed = 4·5ms^-1
Final Speed = 29ms^-1
Time Taken = ?
Acceleration = 10ms^-2

The formula we need to use is the first one.

29 = 4·5 + 10×t
24·5 = 10t
t = 2·5s (2 s.f.)

That's about as easy these questions can get.

Now it's time to get excited, because it turns out that a hollywood film is being filmed a 1km up the road. Lets see what's going on.

The stuntman will accelerate (at ·45ms^-2) from rest towards another cliff 1km away, then jump out of the car before it gets thrown off the cliff. The whole take finishes when the car hits the ground, 90 seconds later. So how fast will the car be going as it goes off the cliff? And how high is the cliff?

Well, the scene isn't the only exciting thing in this example, we get to use three different formulas!

Got a little too excited there, but let's get started with the answer.

First of all, we need to find the speed the car is going before it drives off the cliff.

Initial Speed = 0ms^-1
Final Speed = ?
Distance = 1000m
Acceleration = ·45ms^-2

The only formula that fits our needs is the third formula, so using that it gives us...

v_F² = v_I² + 2ad
v_F² = 0² + 2×·45×1000
v_F² = 900
v_F = 30ms^-1

So that's our first answer. We also need to know how long that part of the shot took, so we can subtract it from the total scene time of 90 seconds. For this we will use the first formula with the information given below.

Initial Speed = 0ms^-1
Final Speed = 30ms^-1
Acceleration = ·45ms^-2
Time = ?

v_F = v_I + at
30 = 0 + ·45t
t = 30÷·45
t = 66.67s

Here is a diagram showing the acceleration due to gravity on the car as it goes off the cliff.

So high is the cliff? For this we need the vertical speed. It does not require the fact that the car is already going 30ms^-1 because the assumption of no air resistance means the car will continue to move horizontally until it hits the ground; so it is irrelevant.

Instead we need to know that its beginning vertical speed is 0, because it was not "thrown" downwards when it went off the cliff.

Initial Vertical Speed = 0ms^-1
Distance = ?
Acceleration = 10ms^-2
Time Taken = 90-66·67 = 23·33s

And using the second formula we get:

d = 0×23·33 + ·5×10×23·33²
d = 2700m (2 s.f.)

There's the second answer! Give yourself a pat on the back, hot chocolate or whatever else is relevant.