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## amishra Group Title 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? one year ago one year ago

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1. imron07 Group Title

$t=\sqrt{\frac{2h}{g}}$ 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. amishra Group Title

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

3. imron07 Group Title

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

4. amishra Group Title

Yes :) Okay, now what?

5. imron07 Group Title

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

6. amishra Group Title

g is what?

7. imron07 Group Title

gravity acceleration, with value 9.8 m/s^2

8. amishra Group Title

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

9. imron07 Group Title

Good, that's correct.

10. amishra Group Title

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

11. imron07 Group Title

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

12. amishra Group Title

Wait, what is v0x and x? :/

13. imron07 Group Title

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

14. amishra Group Title

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

15. imron07 Group Title

Yes, now you can find x.

16. amishra Group Title

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

17. imron07 Group Title

It's okay. I love teaching :D

18. amishra Group Title

And you're damn good at it too :)

19. imron07 Group Title

Lol, thx for that :)