Question

A rock thrown from the top of a vertical cliff with horizontal speed of 12.5 m/s strikes the sea 75m away from the foot :

Answer 1

Answer:

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Answer 2

Answer:

ignoring wind resistance the stone lands 47m away from the base of the cliff

Explanation:

Given: A rock thrown from the top of a vertical cliff with horizontal speed of 12.5 m/s strikes the sea 75m away from the foot

To find:  The horizontal range of an object thrown from a cliff

Solution:

The formula for finding the horizontal range of an object thrown from a cliff is

$Range= initial\ velocity\ in\ the\ horizontal\ direction*time$

or

$R= u(x)*T$

where

R=range=unknown

u(x)= initial velocity along the horizontal or the x-axis=10.0m/s

T= time which is unknown the formula for finding the time is

T=square root of (2*height/acceleration due to gravity)

or

$T=\sqrt{2h/g}$

$T=\sqrt{(2*108.4)/9.8}$

$T=\sqrt{(216.8/9.8)}$

$T=\sqrt{22.122}$

$T=4.7s$

$R=10.0*4.7=47m$

ignoring wind resistance the stone lands 47m away from the base of the cliff