Question

A projectile projected from the ground has its
direction of motion making an angle π/4 with the
horizontal at a height 40 m. Its initial velocity of
projection is 50 m/s, the angle of projection is​

Answer 1

Let a particle is projected at an angle α with horizontal with speed 50 m/s.

at a height of 40m, direction of motion of particle making an angle π/4 or 45° with the horizontal.

we know, velocity of particle along horizontal direction remains constant.

so, 50cosα = vcos45°

or, v = 50√2cosα......(1)

now applying equation, $v_y^2=u_y^2+2a_yy$

or, (vsin45°)² = (usinα)² + 2(-g)h

or, 2500cos²α = 2500sin²α - 2 × 10 × 40

or, 25cos²α = 25sin²α - 8

or, 25 = 25tan²α - 8sec²α

or, 25 = 25tan²α - 8(1 + tan²α)

or, 25 = 17tan²α - 8

or, tan²α = 33/17

or, tanα = ±√(33/17)

or, α = tan^-1(√33/17)) or, tan^-1(-√(33/17))

Answer 2

Here your answer goes↓