(a) The launch angle of the ball is 5⁰.
(b) The actual speed of the ball when it leaves the person's hand is 35.13 m/s.
(c) The maximum height reached by the ball is 0.46 m.
(d) The time spent in air by the ball is 0.26 s.
(e) The horizontal distance traveled by the ball on the road is 4.55 m.
Launch angle of the ball
The launch angle of the ball is calculated as follows;
[tex]tan(\theta) = \frac{V_y}{V_x} \\\\\theta = tan^{-1} (\frac{3}{35} )\\\\\theta = 5\ ^0[/tex]
Actual speed of the ball
The actual speed of the ball when it leaves the person's hand is calculated as follows;
[tex]V = \sqrt{V_x^2 + V_y^2} \\\\V = \sqrt{(35)^2 + (3)^2} \\\\V = 35.13 \ m/s[/tex]
The maximum height reached by the ball
The maximum height reached by the ball is calulated as follows;
[tex]Vy_f^2 = V_y_i^2 - 2gh\\\\0 = 3^2 - 2(9.8)(h)\\\\19.6h = 9\\\\h = \frac{9}{19.6} \\\\h = 0.46 \ m[/tex]
Time spent in air by the ball
The time spent in air by the ball is calculated as follows;
[tex]h=V_yt + \frac{1}{2} gt^2\\\\0.46 = 3t + 4.9t^2\\\\4.9t^2 +3t - 0.46 = 0\\\\a = 4.9, \ b = 3, \ c =- 0.46\\\\t = \frac{-b \ \pm \ \sqrt{b^2 - 4ac} }{2a} \\\\t = 0.13 \ s[/tex]
Total time for upward and downward motion = 2(0.13 s) = 0.26 s
Horizontal distance traveled by the ball
X = Vx(t)
X = 35 x 0.13
X = 4.55 m
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