The range of a projectile fired at an angle of `15^@` is 50 m. If it is fired with the same speed at an angle of `45^@` its range will be

Options

A. 60m
B. 71 m
C. 100 m/s
D. 141 m

Answer

C

Explanation

We know that,
Range ( R) = `(u^(2) sin 2theta)/g`
where `theta` is angle of projection.
Given, `theta = 15^(@)` and R = 50 m
Putting all the given values in the formula, we get-
`rArr R = 50 m = (u^(2) sin (2 xx 15^(@)))/g`
`rArr 50 xx g = u^(2) sin 30^(@) = u^(2) 1/2`
`rArr 50 xx g xx 2 = u^(2)`
`rArr u^(2) = 50 xx 9.8 xx 2 = 100 xx 9.8 = 980`
`rArr 14 xx 2.23 m//s = 31.304 m//s`
For `theta = 45^(@), R = (u^(2) sin 2 xx 45^(@))/g = u^(2)/g (therefore sin 90^(@)=1)`
`rArr R = (14sqrt(5))^(2)/g = (14 xx 14 xx 5)/(9.8) = 100 m`