posted by Chris .
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?
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