Question

Projectile is fired at an angle of 45 with the horizontal. elevation angle of the projectile at its highest point as seen from the point of projection is

Answer 1

The answer is tan inv(1/2).

The height of the projectile is H = u^2 sin^2θ /2g.

H= u^2 sin^2 45/ 2g then H = u^2/4g.

Range of projection (R) = u^2 sin 2θ/g= u^2 sin 90/g= tan α = ½.

The elevation angle of projectile at its higher point as seen from point of projection is ½. α = tan^-1(1/2).

Answer 2

Answer:

Explanation:

Here is the answer