The answer on whether the ball clears the net and by how much is that;
Yes, The ball clears the net by a height of 1.01 m.
We are given;
Initial height of ball; y₀ = 2 m
Initial velocity of ball; v = 20 m/s
Angle above the horizontal; θ = 5°
Horizontal distance to the net; x = 7 m
Height of net; h = 1 m
The formula to find by how much the ball clears the net is by the formula;
h' = y₀ + xtan θ - [(¹/₂gx²)/(v² × (cos θ)²)] - h
Plugging in the relevant values gives;
h' = 2 + 7tan 5 - [(¹/₂ * 9.8 * 7²)/(20² × (cos 5)²)] - 1
⇒ h' = 1 + 0.6124 - (240.1/396.96155)
⇒ h' = 1.6124 - 0.6048
⇒ h' ≈ 1.01 m
In conclusion, the height by which the ball clears the net is 1.01 m
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The ball cleared the net by a horizontal distance of 9.74 m.
The given parameters;
The time of motion of the ball is calculated as follows;
[tex]h = h_0 + v_0t - \frac{1}{2} gt^2\\\\0 = 2 + (20 sin(5))t - (0.5 \times 9.8)t^2\\\\0 = 2 + 1.743t - 4.9t^2\\\\4.9t^2 - 1.743 t - 2 = 0\\\\solve \ the quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ b = -1.743 \ c = -2\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-1.743) \ \ +/- \ \ \sqrt{(1.743)^2 - 4(4.9\times -2)} }{2(4.9)} \\\\t = 0.84 \ s[/tex]
The horizontal distance traveled by the ball is calculated as follows;
[tex]R = v_0 cos(\theta) t\\\\R = 20 \times cos(5) \times 0.84\\\\R = 16.74 \ m[/tex]
The ball cleared the net by (16.74 - 7 )m = 9.74 m
Thus, we can conclude that the ball cleared the net by a horizontal distance of 9.74 m.
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