Question

A particle is projected horizontally with a speed V=5m/s from the top of a plane inclined at an angle `theta=37^(@)` to the horizontal `(g=10m//s^(2))`

How far from the point of projection will the particle strike the plane?
A. 75m
B. `(65)/(16)m`
C. `(75)/(16)m`
D. `(85)/(9)m`

Answer 1

Correct Answer - c
`R=(u cos theta)T+(1)/(2)g sin thetaT^(2)`
`T=(2u sin theta)/(g cos theta)`
`V_(x)= u cos theta+g sin thetaT`