Home
Class 12
PHYSICS
A golfer standing on level ground hits a...

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)`.

Text Solution

Verified by Experts


`v_(y) = sqrt(u_(y)^(2) -2gy)`
`=sqrt(52 xx 52 xx (5 xx 5)/(13 xx 13) -2 xx 10 xx 15)`
`= sqrt(16 xx 25 -300) = 10`
`Deltat =(2u_(y))/10 =(2 xx 10)/10 = 2s`
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise SHORT ANSWER TYPE QUESTIONS|6 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise VERY SHORT ANSWER TYPE QUESTIONS|10 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (Circular Motion)|7 Videos

  • MOTION IN A STRAIGHT LINE

    AAKASH SERIES|Exercise very Short answer type question|15 Videos

Similar Questions

Explore conceptually related problems

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 boy throws a ball with a velocity of 15 m/s at an angle of 15° with the horizontal. The distance at which the ball strikes the ground is

A football player throws a ball with a velocity of 50 m//s at an angle 30 degrees from the horizontal. The ball remains in the air for (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 boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle 30 ^@ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? (g = 10 m//s ^ 2 )

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 cricket ball is thrown at a speed of 30 m s^(-1) in a direction 30^(@) above the horizontal. The time taken by the ball to return to the same level is

A ball is thrown at a speed of 40 m/s at an angle of 60^0 with the horizontal. Find a. the maximum height reached and b. the range of te ball. Take g=10 m/s^2 .

A particle is projected with velocity of 10m/s at an angle of 15° with horizontal.The horizontal range will be (g = 10m/s^2)

A football ils kicked with a velocity of 20 m/s at an angle of 45^0 wilth the horizontal. A. Find the time taken by the ball to strike the ground. b.Find the maximum height it reaches. C. How far away from the kick dows it hit the ground? Take g=10 m/s^2 .