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

A tennis ball is released from rest at a height h above the ground. At each bounce 50% of its K.E. is last to its surroundings. What is the height reacted by the ball after its second bounce?

A
h/8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
h/4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
h/2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Zero
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B h/4
Velocity attained after falling a height h will be v=22gh
first Bounce-
remaining KE=mv212=12×mv22
so velocity in upward direction will be v1=v22
so after the time when it return from its top height it will have same velocity as v1
Now
for second bounce-
incoming KE is mv212
and after the bounce the remaining KE will bemv222=12×mv212
so velocity in upward direction will be v2=v122=v2
so maximum height attained after this bounce will be H=(v2)22g=(v/2)22g=v28g=2gh8g=h4

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Motion Under Gravity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon