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

A ball is thrown up with initial velocity u. The retardation due to the drag is proportional to the velocity with proportionality constant, β. The maximum height and the time taken to reach the maximum height are

A
uβgβ2ln(1+βug),1βln(1+βug)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
uβgβ2ln(1βug),1βln(1βug)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
uβ+gβ2ln(1βug),1βln(1+βug)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
uβgβ2ln(1+βug),1βln(1βug)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A uβgβ2ln(1+βug),1βln(1+βug)
Let the origin be at the ground and vertically up direction be positive. The acceleration of the ball is given by
dvdt=gβv=g(1+γv),......(1)
where γ=βg. Let the ball takes time T to reach the maximum height H.
Initial velocity of the ball is u and its velocity at the maximum height is zero. Integrate equation (1)
0udv1+γv=gT0dt
to get T=1βln(1+βgu).
Re-write dvdt=vdvds in equation (1) and integrate
0uvdv1+γv=gH0ds
To get H=uβgβ2ln(1+βgu)

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