A ball is thrown upward from the top of a 40.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 31.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?
physics - bobpursley, Thursday, September 17, 2009 at 8:39pm
hf=0=40 + 12t-4.9t^2
solve for t, the time when the ball is at the bottom.
avg speed= 31/time