Question on: WAEC Physics - 1991
A ball is projected horizontally from the top of a hill with a velocity of 20m-1. If it reaches the ground 4 seconds later, what is the height of the hill? (g = 10ms-2)
A
20m
B
40m
C
80m
D
160m
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Correct Option: C
Initial horizontal velocity = 20ms-1
Initial vertical velocity = 0
s = ut + \(\frac{1}{2}gt^2\) = 0 x 5 x 42
= 80m
Initial vertical velocity = 0
s = ut + \(\frac{1}{2}gt^2\) = 0 x 5 x 42
= 80m
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