Question

If the time of flight of a bullet over a horizontal range R is T then the angle of projection with horizontal is

Answer 1

If the initial vertical velocity = Uy, and horizontal velocity = Vx... 
θ = arctan(Uy / Vx) 

And Uy = gT / 2 
Vx = R / T 
Hence θ = arctan(gT^2 / [2R] )

Answer 2

Answer:

The angle of projection with horizontal is $\bold{\theta=\arctan \left(\frac{T^{2} g}{2 R}\right)}$

Explanation:

Given

The horizontal range =R

The time taken =T

Let the angle of projection be θ

The initial velocity be u.

Let the vertical component of the velocity will be equal to usinθ

Then when the bullet reaches the maximum height then the velocity will be 0.

$0=\frac{u \sin \theta}{g t}$

Therefore,

Where g=acceleration due to gravity.

t=time taken to attain the maximum height.

Therefore

$t=\frac{u \sin \theta}{g}$

When the bullet moves in the downward then the time taken by the bullet will be the same t at the downward direction.

Hence

$\begin{array}{l}{T=2 t=\frac{2 u sin \theta}{g}} \\ {u=\frac{T_{g}}{2 \sin \theta}}\end{array}$

The horizontal velocity of the bullet= ucosθ

Hence in the time T it would have covered the distance of Tucosθ with the same R.

Therefore

R=T ucosθ

Substitute the value of u in the above eqn

We get

$\begin{aligned} R &=T\left(\frac{T g}{2 \sin \theta}\right) \cos \theta \\ R &=\frac{T^{2} g}{2 \tan \theta} \end{aligned}$

$\tan \theta=\frac{t^{2} g}{2 R}$

$\bold{\theta=\arctan \left(\frac{T^{2} g}{2 R}\right)}$