. A ball is dropped from a height of 90m on a floor. At each collision
with the floor the ball loses one tenth of its speed. Plot the speed time
graph of its motion between t=0 to 12s.
Answers
Answer:
Height from which the ball is dropped s=90m
Initial velocity of the ball, u=0
Let us consider that the final velocity of the ball is v
The time taken by the ball to reach the ground is given by:
s=ut+
2
1
at
2
90=0+
2
1
×9.8t
2
t=
18.38
=4.29s
The final velocity of the ball as it reaches the ground is given by:
v=u+at
⇒v=0+9.8×4.29=42.04m/s
The ball losses it's one-tenth of the velocity at collision.
So, the rebound velocity of the ball,
u
r
=
10
9v
=
10
9×42.04
=37.84m/s
Now, the ball rebounds and reaches a new maximum height from the ground. The time taken by the ball is given by:
v=u
r
+at′
0=37.84+(–9.8)t′
t′=−
−9.8
37.84
=3.86s
Total time the ball takes to reach the maximum height is :
T=t+t′
=4.29+3.86=8.15s
Now the ball travels back to the ground in the same time as it takes to reach the maximum height i.e. 3.86s
The final velocity of the ball with which it reaches the ground will be same as the speed with which it goes up.
So, the velocity of the ball after rebound:
v
′
=
10
9×37.84
=34.05m/s
Total time taken by the ball to rebound for the second time:
T
t
=8.15+3.86=12.01s