A golfer standing on level ground hits a ball with a velocity of `u=50ms^(-1)` at an angle `theta` above the horizontal.If `tan theta=(5)/(12)` then the time for which the ball is at least `15m` above the ground will be take `g=10ms^(-2)`

A golfer standing on level ground hits a ball with a velocity of 52 m s^-1 at an angle theta above the horizontal. If tan theta = 5//12 , then find the time for which then ball is atleast 15 m above the ground (take g = 10 m s^-2) .

A golfer standing on the ground hits a ball with a velocity of 52 m//s at an angle theta above the horizontal if tan theta=5/12 find the time for which the ball is at least 15m above the ground? (g=10m//s^(2))

A golfer standing on level ground hits a ball with a velocity of u=52m//s at an angle alpha above the horizontal. If tantheta=5//12 , then the time interval for which the ball is at least 15 m above the ground (i.e. between A and B will be :(take g=10m//s^(2) )

A ball is thrown from the top of tower with an initial velocity of 10ms^(-1) at an angle of 30^(@) with the horizontal. If it hits the ground of a distance of 17.3m from the back of the tower, the height of the tower is (Take g=10ms^(-2))

A ball is projected with a velocity 20 sqrt(3) ms^(-1) at angle 60^(@) to the horizontal. The time interval after which the velocity vector will make an angle 30^(@) to the horizontal is (Take, g = 10 ms^(-2))

A cricketer hits a ball with a velocity 25m//s at 60^(@) above the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground)

A ball is projected with initial velocity u at an angle theta to the horizontal. Then horizontal displacement covered by ball as it collides third time to the ground would be, if coefficient of restitution is e :

A boy throws a ball with a velocity u at an angle theta with the horizontal. At the same instant he starts running with uniform velocity to catch the ball before it hits the ground. To achieve this he should run with a velocity of