Asked by suvalakshmi
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?
Answers
Answered by
Elena
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
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
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