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

A particle is projected from the ground with an initial speed of v at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

Open in App
Solution



Average velocity = DisplacementTime
=H2+R24T2

we know that, H=v2sin2θ2g, R=v2sin2θg, T=2v sinθg

vav=(gv sinθ)((v2sin2θ2g)2+(v2sin2θ)24g2)
=v4 sin4θ+v4 sin22θ4g2×g2v2sin2θ

On simplifying further,
vav=v21+3 cos2θ

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