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?

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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?

Physics
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\[t=\sqrt{\frac{2h}{g}}\] You can derive this equation if you want. From here you can find h with known t (a). For b, simply: \[x=v_x*t\]
I'm sorry, can you explain it further? I didn't get it :/
Let's start from beginning: \[h=v_{0y}t+\frac12 gt^2\] Familiar with this?

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Yes :) Okay, now what?
since the student throw the stone horizontally, it means v0y=0 can you find h now? :)
g is what?
gravity acceleration, with value 9.8 m/s^2
So, h= (1/2) * 9.8 * (5.5)^2 ?
Good, that's correct.
Oh okay, and how do I find the horizontal distance?
This is the equation for horizontal motion: \[x=v_{0x}t\]
Wait, what is v0x and x? :/
vox is initial horizontal velocity, and x is horizontal displacement.
So, horizontal velocity is 8 m/s? And time is 5.5 seconds?
Yes, now you can find x.
Oh :/ Wow, THANK YOU SO MUCH for spoon feeding me through the entire problem! :D
It's okay. I love teaching :D
And you're damn good at it too :)
Lol, thx for that :)

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