Friday, September 25, 2015

REST AND MOTION Q32

A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground, (c) the velocity (direction and magnitude) with which it strikes the ground.
SOLUTION:=




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
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