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A projectile thrown with an initial spee...

A projectile thrown with an initial speed `u` and the angle of projection `15^@` to the horizontal has a range `R`. If the same projectile is thrown at an angle of `45^@` to the horizontal with speed `2 u`, what will be its range ?

A

`R = 8 R_2`

B

`R_2 = 5 R`

C

`R_2 = R/3`

D

`R_2 = 8 R`

Text Solution

Verified by Experts

The correct Answer is:
D
`R = (u^2 sin 2 theta)/(g) :. R prop u^2 sin 2 theta`
`(R_2)/(R_1) = ((u_2)/(u_1))^2 ((sin 2 theta_2)/(sin 2 theta_1))`
`rArr R_2 = R_1 ((2u)/(u))^2 ((sin 90^@)/(sin 30^@)) = 8 R_1`.
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