Question

A ball is thrown vertically upwards from the top of a building of height 20m with the speed of 20m/sec. (Take 'g'=10m/sec2).Find-velocity at which the ball hits the ground.
-total time of the journey

Answer 1

Question 33179

Let $h$ denote the height of the building. For motion upward until stop, $y = h + v_0 t - \frac{g t^2}{2}$. At the moment of stop, $v = v_0 - g t_2 = 0$, therefore $t_2$ -time of movement until stop at maximum height is $t_2 = \frac{v_0}{g}$. Plugging in this expression into coordinate, obtain $y_2 = h + \frac{v_0^2}{2g}$ - the height at the moment of stop.

Then, motion down is without acceleration. Hence, $v = g t$, and $y = y_{2} - \frac{g t^{2}}{2}$. When ball reaches the ground ($y = 0$), from here obtain the time of movement from highest point to the ground - $t_{f} = \sqrt{\frac{2y_{2}}{g}} = \sqrt{\frac{2}{g}(h + \frac{v_{0}^{2}}{2g})}$. Thus, velocity at which the ball hits the ground is $v = g t_{f} = \sqrt{2g(h + \frac{v_{0}^{2}}{2g})} = 28.28\frac{m}{s}$.

Total time of the journey is $t = t_2 + t_f = \frac{v_0}{g} + \sqrt{\frac{2}{g}(h + \frac{v_0^2}{2g})} \approx 4.83 \, \text{s}$.