A projectile is fired with speed `v_(0)` at `t=0` on a planet named ' Increasing Gravity ' . This planet is strange one, in the sense that the acceleration due to gravity increases linearly with time `t` as` g(t)=bt`, where `b` is a positive constant. 'Increase Gravity'
If angle of projection with horizontal is `theta`, then the maximum height attained is

Options

A. `(1)/(3) ((v_(0)sintheta)^(3//2))/(sqrt(b))`
B. `(4)/(3) ((v_(0)sintheta)^(3//2))/(sqrt(b))`
C. `((2v_(0)sintheta)^(3//2))/(3sqrt(b))`
D. None of these

Answer

C

Explanation

For maximum height `(dy)/(dt)=0=-(dt^(2))/(2)+v_(0)sin theta`
`:. y` is maximum at `t=sqrt((2v_(0)sin theta )/(b))`
or `y_(max)=(-(bt^(2))/(6)+v_(0)sin theta)t`
`=(-(b)/(6)xx(2v_(0)sin theta )/(b)+v_(0)sin theta ) sqrt((2v_(0)sin theta)/(b))`
`=(2)/(3)(v_(0)sintheta)/(sqrt(b))sqrt(2v_(0)sin theta)=((2v_(0)sin theta )^(3//2))/(3sqrt(b))`