A helicopter pilot throws a ball downwards from an unknown height above the ground. The ball has an initial speed of 14.1 m/s. The ball hits the ground 9.2 s later. Find the initial height of the ball above the ground
"Let~v_0=14.1~m\/s,~t=9.2~s,~g=9.81~m\/s^2\\\\\nh=v_0t+gt^2\/2=(v_0+gt\/2)t=\\\\\n=(14.1+9.81\\cdot9.2\/2)\\cdot9.2~(m)\\approx545~(m)"
Comments
Leave a comment