41. A boy standing on a long railroad car throws a ballstraight upward

1.	41. A boy standing on a long railroad car throws a ballstraight upwards. The car is moving on the horizontalroad with an acceleration of 1 m/s and the projectionvelocity in the vertical direction is 9.8 m/s. How farbehind the boy will the ball fall on the car?
Answer» 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

41. A boy standing on a long railroad car throws a ballstraight upwards. The car is moving on the horizontalroad with an acceleration of 1 m/s and the projectionvelocity in the vertical direction is 9.8 m/s. How farbehind the boy will the ball fall on the car?

Answer»

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

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