A student stands at the top of a cliff and throws a stone horizontally at a speed of 8 m/s. They see the stone splash into the water 5.5 seconds later.

a) How high is the cliff?
b) What horizontal distance does the stone travel?

Question

A student stands at the top of a cliff and throws a stone horizontally at a speed of 8 m/s. They see the stone splash into the water 5.5 seconds later. 

a) How high is the cliff?
b) What horizontal distance does the stone travel?

anonymous · Answer

$$t=\sqrt{\frac{2h}{g}}$$
You can derive this equation if you want. From here you can find h with known t (a).
For b, simply:
$$x=v_x*t$$

anonymous · Answer

I'm sorry, can you explain it further? I didn't get it :/

anonymous · Answer

Let's start from beginning:
$$h=v_{0y}t+\frac12 gt^2$$
Familiar with this?

anonymous · Answer

Yes :) Okay, now what?

anonymous · Answer

since the student throw the stone horizontally, it means v0y=0
can you find h now? :)

anonymous · Answer

g is what?

anonymous · Answer

gravity acceleration, with value 9.8 m/s^2

anonymous · Answer

So, h= (1/2) * 9.8 * (5.5)^2  ?

anonymous · Answer

Good, that's correct.

anonymous · Answer

Oh okay, and how do I find the horizontal distance?

anonymous · Answer

This is the equation for horizontal motion:
$$x=v_{0x}t$$

anonymous · Answer

Wait, what is v0x and x? :/

anonymous · Answer

vox is initial horizontal velocity, and x is horizontal displacement.

anonymous · Answer

So, horizontal velocity is 8 m/s? And time is 5.5 seconds?

anonymous · Answer

Yes, now you can find x.

anonymous · Answer

Oh :/
 Wow, THANK YOU SO MUCH for spoon feeding me through the entire problem! :D

anonymous · Answer

It's okay. I love teaching :D

anonymous · Answer

And you're damn good at it too :)

anonymous · Answer

Lol, thx for that :)