Question

A ball is thrown vertically upward from the edge of a 98 m building and reaches the ground 6s after leaving the throwers hand. Assume that the throwers hands is 2 r above the roof of the building calculate the initial velocity of the ball

Answer 1

The total height at which the ball was thrown is H=98+2=100 m. It took the ball some time $t_1$ to go up and reach the highest point of the trajectory above the thrower's hands $h$ and some time $t_2$ to cover the distance $h+H.$ Therefore:

$$t_1+t_2=T,\\\space\\ t_1=\frac vg, \\\space\\ t_2=\sqrt{\frac{2(h+H)}{g}},\\\space\\ h=\frac{v^2}{2g}.\\\space\\ t_2=\sqrt{\frac{2\Big(\frac{v^2}{2g}+H\Big)}{g}},\\\space\\ T=t_1+t_2=\frac vg+\sqrt{\frac{2\Big(\frac{v^2}{2g}+H\Big)}{g}}.\\\space\\ T-\frac vg=\sqrt{\frac{2\Big(\frac{v^2}{2g}+H\Big)}{g}},\\\space\\ \bigg(T-\frac vg\bigg)^2=\frac{2\Big(\frac{v^2}{2g}+H\Big)}{g},\\\space\\ T^2-2\frac{vT}{g}+\frac{v^2}{g^2}=\frac{v^2}{g^2}+\frac{2H}{g},\\\space\\ T^2-2\frac{vT}{g}=\frac{2H}{g},\\\space\\ v=\frac{gT}{2}-\frac{H}{T}=\frac{9.8·6}{2}-\frac{100}{6}=12.7\text{ m/s}.$$