Posted by **Chris** on Thursday, November 15, 2012 at 8:28pm.

A baseball player throws a baseball up at 24.3 m/s. How long must he wait before he can catch the baseball on its way down?

- Physics -
**Jennifer**, Friday, November 16, 2012 at 3:50pm
y = y0 + vy*t + 1/2*ay*t^2

Where y is the distance in the vertical direction, y0 is the initial y position, vy is the speed in the y direction, ay is the acceleration in the y direction, t is time.

Choose a coordinate system so that y0 equals zero at the baseball player's hand. y0=0; vy=24.3; ay = -9.8

0 = 24.3*t - 1/2*9.8*t^2

Dividing by t:

0 = 24.3 - 1/2*9.8*t

Solve for t --this is the time at which the ball reaches 0 again, or the same height from which it was released

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