a train is moving at a speed of 5m/s. a person standing on the train throws a ball upwards at 10m/s. at the maximum height of the ball the train stopped. what is the deceleration of the train and what is the distance between the ball and the person finally?

Answer 1

Assume g=10 m/s².
The ball upward motion
0=v₀-gt,
t=v₀/g=10/10=1 s.
The train motion for time 1 s
0= v-at,
a=v/t=5/1 = 5 m/s²,
the distance
s= v t -at²/2 =5•1 -5•1²/2 = 2.5 m