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Assume the ball's horizontal speed remains constant, since the only significant force is gravity which acts in the vertical direction. We can calculate the time of flight: time = distance / speed = 19.6m / 28ms-1 = 0.7s Consider now the vertical direction: we know that the ball took 0.7s to fall to the ground, and also that its initial vertical speed was zero. Its acceleration was g = 9.81ms-2. It is thus possible to deduce the vertical distance that it fell. s = ut + at2/2 ... but u = 0 = at2/2 = 9.81 x 0.72 / 2 = 2.40m to 3 significant figures.

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Q: A tennis ball is struck such that it leaves the racket horizontally with speed of 28.0 ms the ball hits the court at horizontal distance of 19.6 m from the racket what is the height of the tennis ball?
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