amishra
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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imron07
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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:
\[x=v_x*t\]
amishra
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I'm sorry, can you explain it further? I didn't get it :/
imron07
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Let's start from beginning:
\[h=v_{0y}t+\frac12 gt^2\]
Familiar with this?
amishra
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Yes :) Okay, now what?
imron07
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since the student throw the stone horizontally, it means v0y=0
can you find h now? :)
amishra
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g is what?
imron07
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gravity acceleration, with value 9.8 m/s^2
amishra
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So, h= (1/2) * 9.8 * (5.5)^2 ?
imron07
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Good, that's correct.
amishra
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Oh okay, and how do I find the horizontal distance?
imron07
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This is the equation for horizontal motion:
\[x=v_{0x}t\]
amishra
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Wait, what is v0x and x? :/
imron07
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vox is initial horizontal velocity, and x is horizontal displacement.
amishra
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So, horizontal velocity is 8 m/s? And time is 5.5 seconds?
imron07
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Yes, now you can find x.
amishra
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Oh :/
Wow, THANK YOU SO MUCH for spoon feeding me through the entire problem! :D
imron07
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It's okay. I love teaching :D
amishra
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And you're damn good at it too :)
imron07
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Lol, thx for that :)