Answer :
The horizontal velocity of the ball is 27 m/s
The given parameters;
- horizontal distance, x = 12 m
- height of the net, h = 1 m
The time of motion is calculated as;
h = ut + ¹/₂gt²
1 = 0 + (0.5 x 9.8)t²
1 = 4.9t²
[tex]t^2 = \frac{1}{4.9} \\\\t^2 = 0.205\\\\t = \sqrt{0.205} \\\\t = 0.45 \ s[/tex]
The horizontal velocity of the ball is calculated as;
X = uₓt
[tex]u_x = \frac{X}{t} = \frac{12}{0.45} = 26.7 \ m/s \ \approx 27 \ m/s[/tex]
Thus, the horizontal velocity of the ball is 27 m/s
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