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. 71m
C. 100m
D. 141m

Answer

C

Explanation

We know that, where `theta` is angle of projection

Given, `theta = 15^(@)` and `R = 50 m`
Range, `R = (u^(2)sin 2theta)/(g)`
Putting all the given values in the formula, we get
`rArr R = 50 m = (u^(2) sin (2xx 15^(@)))/(g)`
`rArr 50 xx g = u^(2) sin 30^(@) = u^(2) xx (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 u = sqrt(980) = sqrt(49 xx 20) = 7 xx 2 xx sqrt(5)ms^(-1)`
`14 xx 2.23 ms^(-1) = 31.304 ms^(-1)`
For `theta = 45^(@), R = (u^(2) sin 2 xx 45^(@))/(g) = (u^(2))/(g) ( :' sin 90^(@) = 1)`
`rArr R = ((14sqrt(5))^(2))/(g) = (14 xx 14 xx 5)/(9.8) = 100 m`