Is this solvable? Projectile motion

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SUMMARY

The problem involves calculating the height of a cliff based on the time it takes for a rock thrown horizontally to hit the ground. Using the equation sy = uy*t + 1/2*at^2, where the initial vertical velocity (uy) is zero, the cliff height can be determined. Given that the rock takes approximately 15 seconds to reach the bottom, the calculated height is 1103 meters, assuming no air resistance. The discussion clarifies that the horizontal throw does not affect the time to fall, as both horizontally thrown and dropped objects fall at the same rate.

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amber87
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Homework Statement



You are standing at the top of a cliff and want to know how far it is to the bottom so you throw a rock as hard as you can horizontally over the edge of the cliff and time how long it takes before you hear it hit the bottom. Let us assume it takes around 15 seconds. How high is the cliff?

Homework Equations



sy=uyt+1/2at^2

The Attempt at a Solution



I'm stuck because there's just too many unknowns. I worked out 1103m from the equation above which from memory needs to be divided by sin theta which is also unknown. Can anyone help?
 
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Your answer seems to be correct. The question is a little vague.
I think they mean that you throw the ball horizontally, which means that uy_0 = 0. (that's the initial velocity in the y axis).
That fact that you throw it forwards as hard as you can just enables you to assume that the ball would hit the ground, and not some slope of the cliff.
 
Tomer is correct.Because the ball is thrown horizontally the initial vertical velocity is zero.Looking at it differently if you drop an object from a certain height and at the same time throw a second object horizontally from the same height then both objects take the same time to reach the ground.If the second object is thrown at an angle to the horizontal the times will be different.Of course this is ignoring air resistance.
 

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