Answer:
7.0s
Explanation:
Mass = 0.41kg
F= 81N
t = 0.22s
¤ = 29°
Lo = 86m
From impulse equation,
F*t = m* v
81 * 0.22 = 0.41 * v
Vo = 17.82 / 0.41
Vo = 43.46m/s
Vx= velocity across horizontal plane
Vy = velocity across vertical plane
Vx = Vo * cos ¤
Vy = Vo * sin ¤
Vx = 43.46 * cos 30° = 37.64 m/s
Vy = 43.46 sin 30° = 21.73 m/s
Distance travelled across the vertical plane,
L = Lo + Vy *t + ½gt²
0 = 86 + 21.73t - 4.9t²
4.9t² - 21.73t - 86 = 0
Solving for t in the quadratic equation,
t = 6.96 or -10.04
Using the positive root since time can't be negative, t = 6.96 approximately 7.0s
The time spent in the air by the ball during the motion is 6.9 s.
The given parameters;
The velocity of the ball when hit by the bat is calculated as follows;
[tex]F = ma = \frac{mv}{t} \\\\mv = Ft\\\\v = \frac{Ft}{m} \\\\v = \frac{81 \times 0.22}{0.41} \\\\v = 43.46 \ m/s[/tex]
The time of motion of the ball is calculated as follows;
[tex]h = h_0 + v_y t - \frac{1}{2} gt^2\\\\0 = 86 + (vsin\theta)t - (0.5 \times 9.8 \times t^2)\\\\0 = 86 + (43.46\times sin(29))t \ - \ 4.9t^2\\\\0 = 86 + 21.1 \ t - 4.9 t^2\\\\4.9t^2 - 21.1 t - 86= 0\\\\solve\ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ b = -21.1 , \ c = -86\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-21.1) \ \ +/- \ \ \sqrt{(21.1)^2 - 4(4.9\times -86)} }{2(4.9)}\\\\t = 6.9 \ s[/tex]
Thus, the time spent in the air by the ball is 6.9 s.
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