The time spent in air by the ball is 1.38 s.
The initial horizontal velocity of the ball is 7.25 m/s.
The time spent in air by the ball is calculated by using second kinematic equation as follows;
[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 9.3}{9.8} }\\\\t = 1.38 \ s[/tex]
The initial horizontal velocity of the ball is calculated as follows;
[tex]X = v_0_x t\\\\v_0_x = \frac{X}{t} \\\\v_0_x = \frac{10 \ m}{1.38 \ s} \\\\v_0_x = 7.25 \ m/s[/tex]
A student threw a ball horizontally out of a window 9.3 m above the ground. It was caught by another student who was 10.0 m away.
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