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# A large heavy box is sliding without friction down a smooth inclined plane of inclination theta. from a point P on the bottom of the box  particle is projected inside the box, with speed u(relative to box) at angle alpha with the bottom of the box.(a)Find the distance aling the bottom of the box between the point of projection P and the point Q where the particle lands. The particle does not hit any other surface of the box(b)If horizontal displacement of the particle w.r.t ground is zero.Find the speed of the box w.r.t the ground at the moment when particle was projected.

Chetan Mandayam Nayakar
312 Points
10 years ago

let perpendicular distance between bottom of box be y and let horizontal distance covered in box frame be x, let time duration of flight be

2usin(alpha)/gcos(theta)=t, horizontal displacement w.r.t. box=x=ucos(alpha)t-((gsin(theta)t^2)/2),let initial speed be v, zero displacement condition implies that spped of box w.r.t. INCLINED PLANE is such that x=vt+((gt^2)sin(theta)/2)