WAEC - Physics (1991 - No. 3)
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)
20m
40m
80m
160m
200m
Explanation
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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