Question

The ball is thrown at an angle to the horizon, fell to the ground after 4 seconds. The maximum speed of the ball is twice it’s minimum speed. Find the maximum height of the ball

Answer 1

$$\frac{v_x}{v_0}=\frac{1}{2}=\cos \alpha\\\alpha=60\degree$$

$$T=\frac{2v_0\sin{60}}{g}\\4=\frac{2v_0\sin{60}}{9.8}\\v_0=22.6\frac{m}{s}$$

$$H=\frac{v_0^2\sin^2{60}}{2g}\\ H=\frac{22.6^2\sin^2{60}}{2(9.8)}=19.5\ m$$