Question

Max horizontal range of a projectile motion is 4 km. if the projectile is thrown at an angle of 15 degree

Answer 1

At the risk of answering a homework problem I’ll answer this.

Assumptions: no air friction, fixed muzzle velocity, MKS units

Suppose the angle of firing is a (for angle)

Suppose the muzzle velocity is v (for velocity)

Suppose the range is x (for x coordinate)

The flight time t of the projectile is t = (2 v sin a) / 9.8 because v sin a is the vertical component of the the velocity. Gravity reduces that velocity to 0 in (v sin a) / 9.8 seconds, and it takes another (v sin a)/9.8 seconds to reach the ground.

The distance x = t v cos a because the horizontal component of the velocity is v cos a and it travels for t seconds.

Combining, we have

x = (2 v**2 sin a cos a) / 9.8

So this has an interesting symmetry. There is a second angle b such that cos b sin b = cos a sin a. This is because the sin and the cos can just switch places. In fact, cos b = sin a and sin b = cos a. What is that angle? Left as an exercise for the reader