In a film stunt, a car is driven off a cliff with a horizontal velocity of 12.9 ms-1. The cliff face is vertical, and the cliff is 20.7 m high. Calculate the distance in m from the base of the cliff to the point where the car strikes the ground.(adsbygoogle = window.adsbygoogle || []).push({});

should i be considering this problem in x and y - components of the motion separately.

taking upwards as +ve direction for vectors

displacment = -20.7

initial displacment = 0

inital speed = 12.9

a=-9.8

using it in s = ut + 1/2at^2 i get

20.7 + 12.9t - 4.9t^2

t = 3.757 +ve value taken

s=ut

12.9*3.757= 48.5m as the displacement

and then 48.5 + 20.7 = 69.2m for the distance.

help would be apprecated

thanks in advance

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# Homework Help: Film stunt physics car problem

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