Question

The speed of a projectile when it reaches its maximum height is 0.58 times its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

Answer 1

Speed of the projectile at its maximum height is only along horizontal direction

so at highest point

$v_1 = v_x$

now when he is at half of the maximum height the speed will be in x and y direction both

$v_2 = \sqrt{v_y^2 + v_x^2}$

here it is given that

$v_1 = 0.58 v_2$

$v_x = 0.58\sqrt{v_x^2 + v_y^2}$

$2.97 v_x^2 = v_x^2 + v_y^2$

$1.97 v_x^2 = v_y^2$

also we know that

$v_y^2 = v_{iy}^2 - 2 g \frac{H}{2}$

here we know that maximum height is given as

$H = \frac{v_{iy}^2}{2g}$

$v_y^2 = v_{iy}^2 - 2 g\frac{v_{iy}^2}{4g}$

$v_y^2 = \frac{v_{iy}^2}{2}$

now from above

$1.97 v_x^2 = \frac{v_{iy}^2}{2}$

$1.98 v_x = v_{iy}$

also we know that angle of projection is

$tan\theta = \frac{v_{iy}}{v_x}$

$tan\theta = \frac{1.98v_x}{v_x}$

so angle is

$\theta = tan^{-1} 1.98$

$\theta = 63.3 degree$