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I have a physics problem that says

A 76.0-kg boulder is rolling horizontally at the top of a vertical cliff that is 20m above the surface of a lake. The top of the vertical face of a dam is located 100m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25m below the top of the dam. What must be the minimum speed of the rock just as it leaves the cliff so it will travel to the plain without striking the dam?

I tried to solve the problem using projectile motion formulas. What I did was finding velocity by finding the x and y components. But this problem didn't mention any angle or time, which the formulas required. So I tried to find the travel time of stone but it was not good way to solve this question.

How can I find the minimum speed without using time?

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I tried to edit your question to improve the wording a bit. Please check and see whether the changes are accurate. It would also help if you edit the full details of what you did into the question. – David Zaslavsky♦ Feb 6 '12 at 22:18
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The angle is 0 deg.. – Manishearth♦ Feb 7 '12 at 0:53

closed as too localized by Qmechanic♦ Feb 15 at 15:36

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.

2 Answers

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You know the time to fall a certain distance vertically. How fast horizontally does it have to be moving to move (100m?) horizontally in this time

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Additional free hint: There are two ways of solving these kind of problems. One of them is applying a force balance. What is the other approach? – Bernhard Feb 7 '12 at 7:14
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I think everyone's comment helps my idea!

I solved this question with finding 'time' at first because we have

$y= ?$ $Vf=0$ $Vi=?$ $a=-g$

I used y=$Vi*t-1/2gt^{2}$ and $Vf=Vi-gt$

I got $Vi = gt$ and the height of original position was $20+25$

I got $45 = gt^{2}-1/2gt^{2}$ so $Time = (2*45/g)^{1/2} = 3.03(sec) $

now I applied this to the middle point where object traveled 100m. ($x=100$)

I used $x=Vit$ to find the speed. $100 = Vi*3.03$

$Vi = 100/3.03 = 49.5(m/s)$

I just wondering if the $Vi$ is really minimum velocity or not.
I think it right. but just in case if I make some mistake, could you post about it ?

Thanks !!

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