Question

a lemming running 2.87 m/s runs off a horizontal cliff. It lands in the water 5.32 m from the base of the cliff. How high was the cliff?

Answer 1

Looking at the scenario, you can tell that this is a Type 1 Projectile, or a horizontal projectile. You would need a couple of things before you can solve this problem.

First you need to take a look at your given:

Vix = 2.87 m/s (Horizontal velocity)
dx = 5.32m (Horizontal distance)

What you need is your vertical distance or your height. The standard formula for projectile displacement is:
$d = vit + \frac{1}{2}gt^{2}$

Where:
vi = initial velocity
g = Acceleration due to gravity = 9.8m/s^2
t = time

To specifically get the vertical distance you just need to put in the y-components. Now remember that initially, for a horizontal projectile, there is not vertical movement, only forward. So your vi = om/s which will leave you with the equation:

$dy=viyt + \frac{1}{2}gt^{2}$ 
$dy=(0m/s)t + \frac{1}{2}gt^{2}$ 
$dy=\frac{1}{2}gt^{2}$ 

Your new formula would then be:
$dy=\frac{1}{2}gt^{2}$ 

But if you look at your problem, you can see that there is no time given. With what was given to you, you can solve it by deriving it from the x component formula:

$dx=vixt$

Plug in what you know and solve for what you don't know. 
$dx=vixt$
$5.32m=(2.87m/s)t$
$\frac{5.32m}{2.87m/s}=t$
$1.85s=t$

Your time is then 1.85s.

Now that you know the time, you can now solve for your vertical distance/displacement:

$dy=\frac{1}{2}gt^{2}$
$dy=\frac{1}{2}(9.8m/s^{2}(1.85s)^{2}$ 
$dy=\frac{1}{2}(9.8m/s^{2}(1.85s)^{2}$ 
$dy=16.77m$ 
 

The cliff was then 16.77m high.