A projectole fired with initial velocity u at some angle `theta` has a range `R`. If the initial velocity be doubled at the same angle of projection, then the range will be

Options

A. 20 m
B. 40 m
C. 120m
D. 160 m

Answer

D

Explanation

According to given condition,
`(h_(1))/(h_(2)) = (sin^(2)theta)/(sin^(2)(90^(@)-theta)) = (20)/(80)`
`rArr tan^(2)theta = (1)/(4) rArr tan theta = (1)/(2)`
`sin theta = (1)/(sqrt(5))` and `cos theta = (2)/(sqrt(5))`
So, `h = (u^(2)sin^(2)theta)/(2g)`
`rArr 20 = (u^(2))/(10g)` or `(u^(2))/(g) = 200`
`:.` The range, `R = (u^(2) xx 2 sin theta cos theta)/(g) = (200 xx 2 xx 2)/(sqrt(5) xx sqrt(5))`
`= (200 xx 4)/(5) = 160 m`