Question

At what angle to the horizon was the stone thrown if the ratio of its potential energy to the kinetic energy in the upper point of the
trajectory is equal to 0, 2. The answer should be expressed in degrees. Consider the force of air resistance to be negligible.

Answer 1

In the upper point, the velocity equals

$$v_{up}=v_x=v\text{ cos}\alpha.$$

The kinetic energy is

$$KE=\frac12 mv_x^2=\frac12 mv^2\text{ cos}^2\alpha.$$

The potential energy is

$$PE=mgh,$$

where the maximum height is

$$h=\frac{v^2\text{sin}^2\alpha}{2g},\\\space\\ PE=\frac{mv^2\text{sin}^2\alpha}{2}.$$

The ration of the energies is 0.2:

$$0.2=\frac{PE}{KE}=\frac{\text{sin}^2\alpha}{\text{cos}^2\alpha}=\text{tan}^2\alpha,\\\space\\ \alpha=\text{atan}\sqrt{0.2}=24°.$$