Calculating Height of Racket Ball Leaving Racket

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The discussion focuses on calculating the height of a racket ball when it leaves the racket, given its horizontal speed of 4.81 m/s and a horizontal distance of 1.97 m before hitting the court. The user correctly determines the time of flight using the formula x = vt, resulting in t = 0.40956 seconds. To find the height, the user proposes using the equation y = gt²/2, where g represents the acceleration due to gravity, confirming a logical approach to the problem.

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A racket ball is struck in such a way that it leaves the racket with a speed of 4.81 m/s in the horizontal direction. When the ball hits the court, it is a horizontal distance of 1.97 m from the racket.
Find the height of the racket ball when it left the racket.

okay so what I did was try to find the time so I did x=volt which is 1.97=4.81t and I got t=.40956... I think I am on the right path so to find height now can I just do y=gt^2/2 to get my height?
 
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That should work.
 

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