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

Horizontal range equals:

$$R = v \cos \alpha T$$

$\alpha$ – angle of projection with horizontal.

And time of flight of a bullet equals:

$$2 v \sin \alpha = g T$$

From the first equation we get:

$$v \cos \alpha = \frac {R}{T}$$

From the second:

$$v \sin \alpha = \frac {g T}{2}$$

Therefore, divide (2)/(1):

$$\tan \alpha = \frac {g T ^ {2}}{2 R}$$

Or:

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

Answer: $\alpha = \arctan \left(\frac{gT^2}{2R}\right)$