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. 60 m
B. 71 m
C. 100 m
D. 141 m

Answer

C

Explanation

We know that
where `theta` is angle of projection
Given, `" " theta=15^(@)` and R=50m
`"Range", R=(u^(2)sin 2 theta)/(g)`
Putting all the given values in the formula, we get
`rArr " " R=50m=(u^(2)sin (2xx15^(@)))/(g)`
`rArr " " 50xxg = u^(2) sin 30^(@) =u^(2)xx(1)/(2)`
`rArr " " 50xxgxx2xxu^(2)`
`rArr " " u^(2)=50xx9.8xx2=100xx9.8=980`
`rArr " " u=sqrt(980)=sqrt(49xx20)=7xx2xxsqrt(5)m//s`
`=14xx2.23m//s=31.304 m//s`
For `" " theta =45^(@), R=(u^(2)sin 2 xx45^(@))/(g)=(u^(2))/(g) " " ` (`:. sin 90^(@)=1`)
`rArr " " R=((14sqrt(5))^(2))/(g)=(14xx14xx5)/(9.8)=100m`