Question

A particle is projected from a point at the foot of a fixed plane, inclined at an
angle of 45° to the horizontal, in the vertical plane containing the line of
greatest slope through the point. If the particle strikes the plane horizontally
and 0 (> 45°) is the angle of launch measured to the horizontal, then
find the value of tan 0.​

Answer 1

Answer:

Let the particle be projected from O with velocity u and strike the plane at a point P after time t.

Let ON=PN=h, then OP=h2–√.

If the particle strikes the plane horizontally, then its vertical component of velocity at P at zero. Along horizontal direction

h=(ucosϕ)(t)....(1)

Along vertical direction, 0=usinϕ−gt

or usinϕ=gt...(2)

and h=usinϕt−12gt2 ---(3)

Using eqns.(1) and (2) in (3)

(ucosϕ)(t)=(ucosϕ)(t)−12(ucosϕ)(t)tanϕ=2