Question

A projectile is at an angle “θ” from the horizontal at the speed “u”. If
an acceleration of “g/2” is applied to the projectile due to wind in
horizontal direction, then find the new time of flight, maximum
height and horizontal range.

Answer 1

Answer:

ANSWER

Let the angle of projection be θ and speed of projectile be u.

Initially, horizontal component of velocity  ux=ucosθ

Acceleration in horizontal direction  a=2g

So, horizontal range of projectile   R′=uxT+21aT2

where T=g2usinθ is the time of flight.

Or    R′=(ucosθ)g2usinθ+4g×(g2usinθ)2

Or    R′=gu2sin2θ+22gu2sin2θ

Using  R=gu2sin2θ   and   H=2gu2sin2θ

⟹ R′=R+2H

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