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A ball is thrown horizontally from a cliff at a speed of 15 m/s and strikes the ground 45 meters from the base of the cliff. How high was the cliff?

1 Answer
Nov 19, 2017


Assuming air resistance is negligible, we can state that the ball flew for 3 seconds (#t = d/v#)


This then allows us to use a kinematic equation (SUVAT equation) to solve for the vertical distance travelled. Initial (vertical) velocity, called ‘u’ below is zero as the ball is thrown horizontally.

We’ll use #s = u.t + 1/2 a.t^2# knowing the acceleration to be 9.81 #m/s^2# (I guess the ball hits the earth, not another body!) and that u in the vertical direction is zero, we get #s = 1/2a.t^2#

#s = 1/2xx9.81(3^2)#

#s = 44.145# m

I’d quote this to 2 sig. figs., as the data in the question implies.