Question

A bomber is moving with a velocity v (m/s) above H meter from ground. The bomber released a bomb to hit a target T when the sighting angle is tetha. Then the relation between tetha, H and v is - ?​

Answer 1

Hello Dear,

◆ Answer -

tanθ = √H / v√(10g)

◆ Explaination -

When bomb is dropped, it will have zero vertical velocity and horizontal velocity be v.

Let t be time time taken by bomb to reach ground.

H = u.t + 1/2 gt^2

H = 0×t + 1/2 gt^2

t = √(10gH)

Horizontal distance traveled by bomb is -

s = v.t

s = v√(10gH)

When sighting angle is

θ, tanθ is given by -

tanθ = H/s

tanθ = H / v√(10gH)

tanθ = √H / v√(10g)

Thanks dear. Hope this helps you..

Answer 2

Answer:

hope it helps you it is the half answer