A ball is projected form ground with a speed of `20 ms^(-1)

Options

A. 5m
B. 7.5m
C. 10m
D. 12.5m

Answer

B

Explanation

Since, `u sin theta xx t = 10 m`
`rArr 20 sin 45^(@) t = 10`
`rArr t = (10)/(20 sin 45^(@)) = (1)/(sqrt(2))s`
Now, `y = (20 sin 45^(@)) t - (1)/(2) g t^(2)`
`= 20 xx (1)/(sqrt(2)) xx (1)/(sqrt(2)) - (1)/(2) xx 10 xx (1)/(2) = 7.5m`