Home
Class 11
PHYSICS
A golfer standing on the ground hits a b...

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))`

Text Solution

Verified by Experts

The correct Answer is:
`(2)`
`y = u sin theta t - (1)/(2) "gt"^2`
`15 = 52 xx (5)/(13) t - (1)/(2) xx 10 t^2`
`5t^2 + -20t + 15 = 0`
`t^2 - 4t + 3 = 0`
`t^2 - 3t xx t + 3 = 0 rArr t (t - 3)-1(t - 3) = 0`
Thus, `t_2 - t_1 = 3 - 1 = 2 s`.
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS-2

    CENGAGE PHYSICS|Exercise Exercise Comprehension|21 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Integer|9 Videos

  • KINETIC THEORY OF GASES

    CENGAGE PHYSICS|Exercise Compression|2 Videos

Similar Questions

Explore conceptually related problems

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 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 body is thrown with a velocity of 9.8 m//s making an angle of 30^(@) with the horizontal. It will hit the ground after a time

A ball is thrown from the top of a tower with an intial velocity of 10 m//s at an angle 37^(@) above the horizontal, hits the ground at a distance 16 m from the base of tower. Calculate height of tower. [g=10 m//s^(2)]

a ball is thrown with 25 m//s at an angle 53^(@) above the horizontal. Find its time to fight, maximum height and range.

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 thrown from the top of a tower with an initial velocity of 10 m//s at an angle of 30^(@) above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. The height of the tower (g=10m//s^(2)) will be

A stone is thrown up from the top of a tower 20,m with a velocity of 24m/s a tan elevation of 30^(@) above the horizontal. Find the horizontal distance from the foot of the tower to the point at which the stone hits the ground. Take g = 10 m//s^(2)

A ball is thrown from the ground with a velocity of 20sqrt3 m/s making an angle of 60^@ with the horizontal. The ball will be at a height of 40 m from the ground after a time t equal to (g=10 ms^(-2))

A balloon is going vertically upwards with a velocity of 10 m//s . When it is 75 m above the ground, a stone is gently relesed from it. The time taken by the stone to reach the ground is (g=10 m//s^(2))