The ceiling of a hall is 40m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of `56ms^(-1)` without the ceiling of the hall is

Options

A. `25^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

b

Explanation

Here, `u=56ms^(-1)`
Let `theta` be the angle of projection with the horizontal have maximum range, with maximum height 40m
Maximum height, `H=(u^(2)sin^(2)theta)/(2g)`
`40=((56)^(2)sin^(2) theta)/(2xx9.8)`
`sin^(2) theta =(2xx9.8xx40)/((56)^(2))=1/4 or sin theta=1/2`
`theta=sin^(-1)(1/2)=30^(@)`