Question

What is the velocity of projection and angle of projection of a projectile whose maximum height is 200 m and velocity at that instant is 45 m/s?

Please give answer in detail...​

Answer 1

Explanation:

When a projectile reaches maximum height, the vertical component of its velocity is momentarily zero (vy = 0 m/s). However, the horizontal component of its velocity is not zero.

Answer 2

Explanation:

Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u .

The point O is called the point of projection, θ is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight.

Now, 

(a). The total time of flight is

 Resultant displacement is zero in Vertical direction.

Therefore, by using equation of motion

  s=ut−21gt2

 gt=2sinθ

 t=g2sinθ

(b). The  horizontal range is

Horizontal range OA = horizontal component of velocity × total flight time

  R=ucosθ×g2usinθ

 R=gu2sin2θ

(c). The maximum height is

It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero.

By using equation of motion

  v2=u2−2as

 0=u2sin2θ−2gH

 H=2gu2sin2θ

Hence, this is the required solution