1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A particle is projected horizontally with a speed u from the top of a plane inclined at an angle θ with the horizontal. How far from the point of projection will the particle strike the plane?

A
2u2gtanθsecθ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2u8tan2θsecθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2u2gtanθcosθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2ugtanθcos2θ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 2u2gtanθsecθThe horizontal distance covered by the projectile in time t is utThe vertical distance covered during that time is gt22tanθ = vertical distancehorizontal distance = gt22ut=gt2ut=2u tanθgHorizontal distance = 2u2tanθg , vertical = 2u2tan2θgSo distance between point and point of projection = √(2u2tanθg)2+(2u2tan2θg)2=2u2tanθ secθg

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos