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A projectile has initially the same hori...

A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of `3 m s^-2` for `0.5 min`. If the maximum height reached by it is `80 m`, then the angle of projection is `(g = 10 ms^-2)`.

A

`tan^(-1) (3)`

B

`tan^(-1) ((3)/(2))`

C

`tan^(-1)((4)/(9))`

D

`sin^(-1) ((4)/(9))`

Text Solution

Verified by Experts

The correct Answer is:
B
`R_(x)=v sqrt((2h)/(g))hati, R_(y)=v sqrt((2h)/(g))hatj, R= sqrt(R_(x)^(2)+R_(y)^(2))`
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