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

From an elevated point A a stone is projected verticalyy upwards. When the stone reaches a distance h below A its velocity is double of what it was at a height above A. The greatest height attained by the stone is nh/3. What is the value of n?

Open in App
Solution

Given :
Distance=h
Initial velocity of stone =u

Velocity at point B , "h" above A is :
Vb²=u²-2gh
Vb=√(u2-2gh)

velocity at point C below "h"
Vc²=u2-2g(-h)

vc²=u2+2ghvc
=√(u2+2gh)

But according to given :

Vc²=4Vb²

(√(u1+2gh)²=4√(u2-2gh)²

u²+2gh=4u²-8gh

u²=10gh/3

Maximum height attained by stone:

Hmax=u²/2g

=10gh /3x2g=5h/3
Hence the value of n=5

flag
Suggest Corrections
thumbs-up
21
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon