A tennis ball is released from rest at a height h = 32 m above the ground.
The ball collides with the ground and bounces up at 75 % of the impact
speed the ball had with the ground. The collision with the ground is
nearly instantaneously. A second ball is released above the first ball from
the same height the instant the first ball loses contact with the ground.
Calculate the time when the two balls collide in seconds.
(Take g = 9.8 m/s
2
)
i need exact answer
Answers
Answer:
Given that,
Height = h
Final velocity = v
Initial velocity = u
Coefficient restitution = e
We know that,
Restitution = the square root of the ratio of the kinetic energy before and after collision
e=
u
v
=
K
0
K
1
K
1
=e
2
K
0
K
2
=e
2
K
1
K
2
=e
4
K
0
Similarly,
K
3
=e
4
K
2
K
3
=e
6
K
0
We know that,
The kinetic energy is
mgh=K
h=
mg
K
The ball is initially dropped from the height hand acquires a kinetic energy K
0
when it hits the ground, before the first collision.
mgh=K
0
Now, Ifh
1
, h
2
and h
3
are the heights after the first, second and third collisions
h
1
=
mg
K
1
=e
2
mg
K
0
=e
2
h
h
2
=e
4
h
h
3
=e
6
h
Hence, the height raised by the ball after third collision is e
6
h
Explanation:
hope it is helpful for you
Answer:
Let the two balls collide t sec after the first ball is thrown, and let h be the height at which they collide.
For the first ball:
h=40t−
2
1
gt
2
And for the second ball:
h=40(t−2)−
2
1
g(t−2)
2
40t−
2
1
gt
2
=40(t−2)−
2
1
g(t−2)
2
2gt=80+20=100
t=5 sec
∴h=40×5−
2
1
×10×5
2
=75m