wiz-icon
MyQuestionIcon
MyQuestionIcon
6
You visited us 6 times! Enjoying our articles? Unlock Full Access!
Question

A ball is droped from a high rise platform t=0 starting from rest. After 6 s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s. What is the value of v?

A
75 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
64 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
84 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
94 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 75 m/s
The distance traveled by first ball in t=18s is h1=ut+12gt2=0(18)+0.5(10)(18)2=1620
As the second ball travels same distance h1 in time t=(186)=12s, so
1620=vt+0.5gt2 or 1620=12v+0.5(10)(12)2 or 12v=900
So, v=75m/s

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relative Motion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon