Question

A projectile strikes the inclined plane at an angle phi -perpendicular -normal

Answer 1

Answer:

R={2Vo^2}/{g} x {sinФ cosФ(Ф +∅) / {cos^2Ф }

Explanation:

Let the particle strike the plane at end point so that R is the range of the projectile on inclined plane.

Here we are given that the initial speed of particle is Vo

Angle of particle from incline surface = θ

Angle of incline surface from horizontal =Φ

The formula for range R given as

If particle is moving upward

R={2Vo^2}/{g} x {sinФ cosФ(Ф +∅) / {cos^2Ф }

Answer 2

Answer :

B

Solution :

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