The range of a projectile launched at an angle of `15^(@)` with horizontal is 1.5km. The range of projectile when launched at an angle of `45^(@)` to the horizontal is

Options

A. 1.5km
B. 3km
C. 6km
D. 0.75km

Answer

b

Explanation

`R_(1)=(u^(2)sin(2xx15^(@)))/(R)=(u^(2))/(2g)`
`R_(2)=(u^(2))/(g)sin(2xx45^(@))=(u^(2))/(g)`
`therefore R_(2)=2R_(1)=2xx1.5=3.0km`