Question

A projectile is thrown from ground with velocity 50 m per second towards an inclined plane which is at some distance from the point of projection such that it strikes the inclined plane perpendicularly the angle of projection from the ground 53 degree with the horizontal and inclined plane is that the angle of 45 degree to the horizontal then the time of flight is

Answer 1

lets assume the particle is thrown from a vertical height=h, since it is thrown horizontally, its vertical initial vel is zero. so time taken to cover h distance t=2hg−−√

now horizontal distance x=h=50t (h/x=tan45=1)

h=x=500

so range in inclined plane is =5002–√

i neglected the possibility of the particle rolling down which would make the qustn invalid