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A boy standing on a long railroad car throws a ball staight upwards.The car is moving on the horizontal road with an acceleration of 1m/s 2 and the projection velocity in the vertical direction is 9.8m/s. How far behind the boy will the ball fall on the car?
hi, till the time , ball is in the hand of bo , it will have same velocity & accl. of car. & after it is thrown , it will continue in the horizontal velocity with accl. zero , which it has just before leaving the car let , the velocity of car at the moment of throwing of ball = u time taken by ball to come back again on the ground = 2(vertical velo.)/g = 2 sec during this time distance travelled by ball, Sb = ut during this time distance travelled by the car Sc = ut + 1/2 a t^2 it shows that ball will fall at a distance 1/2 a t^2 behind the boy 1/2 a t^2 = 1/2 *1 * (2)^2 = 2 m.
hi,
till the time , ball is in the hand of bo , it will have same velocity & accl. of car. & after it is thrown , it will continue in the horizontal velocity with accl. zero , which it has just before leaving the car
let , the velocity of car at the moment of throwing of ball = u
time taken by ball to come back again on the ground = 2(vertical velo.)/g
= 2 sec
during this time distance travelled by ball, Sb = ut
during this time distance travelled by the car Sc = ut + 1/2 a t^2
it shows that ball will fall at a distance 1/2 a t^2 behind the boy
1/2 a t^2 = 1/2 *1 * (2)^2
= 2 m.
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