wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A ball initially at rest, is dropped from building of height 10 m to ground as shown in the figure below. If the coefficient of restitution is e=0.7 and after collision, the ball rebounds back up to height h, find the value of h. (Take g=9.8 m/s2)

A
10 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4.9 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
9.8 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 4.9 m
Taking downward direction as +ve
Just before collision with ground
Velocity of the ball (u1)=2gh=2×9.8×10=14 m/s

After collision,
Let velocity of the ball (v1)=v m/s
As ground is stationary
Velocity of the ground before collision u2 and after collision v2 will be 0 m/s.

From, coefficient of restiution (e)=v2v1u1u2
0.7=0(v)140
v=14×0.7=9.8 m/s

After rebound, at maximum height h, speed of the ball will become zero.

According to law of conservation of energy,
12mv2=mgh
h=v22g=(9.8)22×9.8
=4.9 m

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon