Question

A 0.5 kg ball is thrown upwards with an initial velocity of 6 m/s. How high will the ball reach before it starts falling again?

Answer 1

Equations of motion of the ball are $v(t) = v_0 - g t$ and $y(t) = v_0 t - \frac{g t^2}{2}$.

When the ball reaches its highest point, the speed is zero, hence $v(t_s) = 0 \Rightarrow t_s = \frac{v_0}{g}$.

$t_s$ is the time it took to reach the highest point. Therefore, $y$ coordinate at that moment of time is how high will the ball reach before it starts falling again: $H = y(t_s) = \frac{v_0^2}{g} - \frac{v_0^2}{2 g} = \frac{v_0^2}{2 g} \approx 1.83 m$.