Question

if the time of flight of a bullet over a horizontal range
Ris T. then the angle of projection with horizontal is
2R?
(1) tan^-1(gT^2/2R)
(2)tan^-1(2R^2/gT)
(3)tan^-1(2R/g^2T)
(4)tan^-1(2R/gT)
​

Answer 1

Answer:

Let the angle of projection of the bullet from the horizontal be θ .

Let the initial velocity be “u”.

The vertical component of this velocity would be usinθ.

When the bullet reaches the maximum height the vertical component of its velocity is “0”.

Therefore,

0= usinθ-gt

Where g is acceleration due to the gravity and “t” is time taken to attain the maximum height.

Therefore

t= usinθ/g

The bullet will take same time “t” in its downward movement as it had taken to go up.

Therefore

T=2t = 2usinθ/g

Or

u= Tg/2sinθ

The horizontal velocity of the bullet is ucosθ.

Therefore in time “T” it would cover a distance of

Tucosθ which is same as “R”

Therefore

R= Tucosθ

Using the value of u as determined above,

R= T x (Tg/2sinθ) x cosθ

R= T^2 g /2tanθ

tanθ= T^2g/2R

θ= arctan ( T^2g/2R)