Question

A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s² and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

Answer 1

we shall use the following kinematic equation to describe the vertical motion of the ball to to its maximum height (half its journey in flight)

v = u + at

here

v = o

u = 9.8 m/s

a = -9. m/s²

thus,

0 = 9.8 - 9.8t

so, the time taken by the ball to reach maximum height will be

t = 9.8 / 9.8 = 1 secs.

or total time of flight, T = 2t = 2 secs

now, the horizontal distance travelled by car in 2 seconds will be

s = uT + (1/2)aT2

here,

u = 0

T = 2 secs.

a = -1 m/s2

thus,

s = 0 + (1/2).-1.2²

so, the distance between ball and car will be

s = -2m