1.

A ball Is projected horizontally from the top of a tower with velocity 4m/s. The velocity of the ball after 0.7s (g = 10m/s^2 ) is

Answer»

11m/s
10m/s
8m/s
3m/s

Solution :VELOCITY of the ball after TIME
t = `sqrt(v^2x +v^2y) =sqrt(4^2+ g t^2)= sqrt(16+(10xx7)^2)` = 8m/s.


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