A cricketer hits a ball with a velocity `25m//s` at `60^(@)` above the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground)

Options

A. 8.2 m
B. 9.0 m
C. 11.6 m
D. 12.7 m

Answer

A

Explanation

`v_(x) = 25 cos60^(@) = 12.5 m//s`
`v_(y) = 25 sin60^(@) = 12.5sqrt(3) m//s`
Time to cover 50 m distance
`t = 50/12.5 = 4` sec
The vertical height
`y = v_(y)t -1/2 "gt"^(2) = 12.5sqrt(3) xx 4 - 1/2 xx 9.8 xx 16`
`=8.1 m = 8.2` m