Answer to Question #130850 in Mechanics | Relativity for Jackson

Question

Describe how the time of flight for an object being thrown vertically upwards to a height of ten metres can be determined

Answer 1

When an object is thrown upwards in the gravity field, its acceleration equals $a=-g\approx-10\frac{m}{s^2}$ . Let the initial velocity be $v_0$. The current heigth of an object above the ground can be determined as $h=v_0t+\frac{at^2}{2}=v_0-\frac{gt^2}{2}$, and the current speed is $v=v_0+at=v_0-gt$. In the upper position $(h=10 m)$ , the velocity equals $v=0$, so $0=v_0-gt_1\Rightarrow v_0=gt_1$ (i). For $h=10m$ we have also: $10=v_0t_1-\frac{gt_1^2}{2}$ (ii). Putting (i) into (ii): $10=gt_1^2-\frac{gt_1^2}{2}\Rightarrow gt_1^2=20 \Rightarrow t_1=\sqrt{2}s.$ $t_1$ is the time when the object is located in the upper position. As we know, the time needed for an object to go up equals to its falling time, so the flight time for the given object equals $t_2=2t_1=2\sqrt{2}s$.