## anonymous 3 years ago 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?

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

Yes :) Okay, now what?

5. anonymous

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

6. anonymous

g is what?

7. anonymous

gravity acceleration, with value 9.8 m/s^2

8. anonymous

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

9. anonymous

Good, that's correct.

10. anonymous

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

11. anonymous

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

12. anonymous

Wait, what is v0x and x? :/

13. anonymous

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

14. anonymous

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

15. anonymous

Yes, now you can find x.

16. anonymous

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

17. anonymous

It's okay. I love teaching :D

18. anonymous

And you're damn good at it too :)

19. anonymous

Lol, thx for that :)