Posted by **micole** on Sunday, May 13, 2007 at 11:38pm.

A ball is thrown upward from the top of a 27.0-m-tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 31.2 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?

Calculate the time t it will take the ball to reach the ground.

27 = -12 t + (g/2)t^2

where g = 9.8 m/s^2

Solve the quadratic for t and take the postive root. (There will be two roots)

Once you have t, to catch the ball, require that

31.2 meters = V t

Solve for the velocity V

calulating the quadratic= 3.872 sec. 31.2 m/ 3.872sec= 8.057m/s. Thank you so much, the answer is correct!

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