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

A projectile is fired with speed v_(0) at t=0 on a planet named ' Incr

1.	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
Answer» `(1)/(3) ((v_(0)sintheta)^(3//2))/(SQRT(b))` `(4)/(3) ((v_(0)sintheta)^(3//2))/(sqrt(b))` `((2v_(0)sintheta)^(3//2))/(3sqrt(b))` None of these Solution :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))`