28. A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with
the floor for 0.01 sec, then average acceleration during contact is
(1) 2100 m/s2
(2) 1400 m/s2
(3) 700 m/s2
(4) 400 m/s2
Answers
Answered by
2
Velocity of the particle at t=2 s
v
=
u
+
a
t where t=2 s
∴
v
∣
∣
∣
∣
∣
t=2s
=(2
i
^
+3
j
^
)+(4
i
^
+2
j
^
)2=10
i
^
+7
j
^
m/s
Displacement of particle
S
=
u
t+
2
1
a
t
2
∴
S
∣
∣
∣
∣
∣
t=2s
=(2
i
^
+3
j
^
)×2+
2
1
×(4
i
^
+2
j
^
)×2
2
=12
i
^
+10
j
^
m/s
2
Answered by
2
Answer:
Let u be the velocity with which the ball hits the ground, then
=2gh
=2×9.8×10=196
∴u=14m/sec
If v be the velocity with which it rebounds, then
=2×9.8×2.5=49
⇒v=7m/sec
∴Δv=(v−u)
=(7m/sec)−(−14m/sec)
=21m/sec
∴a= Δt\Δv
=21/0.01
=2100m/s ^2
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