A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up an

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A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball's initial velocity?

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I had a similar problem for my Webassign physics homework, but the accepted answer was different. There, the intended meaning of the time it takes for the ball to pass the window is merely the time the ball passes in front of the window, not the time from the initial throw. To find the solution in this case you have to use equation 3 to solve for the initial velocity the ball has when it reaches the bottom of the window by using the height of the window and the time it takes for the ball to pass the window, then equation 4 to solve for the initial velocity when thrown using the displacement between the bottom of the window and the ground and the velocity of the ball when it reaches the bottom of the window.

williamwesner