Question

A ball is thrown horizontally from a point 100m above the ground with a speed of 20 m/s. What is the velocity (magnitude and direction) with which it strikes the ground?

Answer 1

$u_x=20\ m/s\\ \\after\ falling\ through\ 100m\\ u_y= \sqrt{2gh}= \sqrt{2 \times 9.8 \times 100} =44.27\ m/s\\ \\magnitude,\ u= \sqrt{u_x^2+u_y^2}\\u= \sqrt{20^2+44.27^2}\ m/s\\u= \sqrt{400+1960}\ m/s\\u= \sqrt{2360}\ m/s=48.58\ m/s\\ \\direction, \theta=tan^{-1}( \frac{u_y}{u_x} )\\ \theta=tan^{-1}( \frac{44.27}{20} )\\ \theta=66.69^0$