Explanation:
At the maximum height, the ball's velocity is 0.
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)
x = 7.72 m
The ball reaches a maximum height of 7.72 m.
The times where the ball passes through half that height is:
x = x₀ + v₀ t + ½ at²
(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²
3.86 = 12.3 t - 4.9 t²
4.9 t² - 12.3 t + 3.86 = 0
Using quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8
t = 0.368, 2.14
The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.
The maximum height the ball reaches above where it leaves your hand is 7.72 m
The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.
The given parameters;
initial velocity, u = 12.3 m/s
acceleration due to gravity, g = 9.8 m/s²
The maximum height the ball reaches above where it leaves your hand is calculated as;
[tex]v_f^2 = v_0^2 - 2gh\\\\at \ maximum \ height \ final \ velocity \ v_f = 0\\\\2gh = v_0 ^2\\\\h = \frac{v_0^2}{2g} \\\\h = \frac{(12.3)^2}{2(9.8)} \\\\h = 7.72 \ m[/tex]
The time for the ball to reach half of the maximum height is calculated as;
[tex]half \ of \ the \ maximum \ height = \frac{7.72}{2} = 3.86 \ m[/tex]
[tex]h = v_0t - \frac{1}{2} gt^2\\\\3.86 = 12.3t - 0.5\times 9.8t^2\\\\3.86 = 12.3t - 4.9t^2\\\\4.9t^2- 12.3t + 3.86=0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ \ b = -12.3, \ \ c = 3.86\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-b \ \ +/- \ \ \sqrt{(-12.3)^2 - 4(4.9\times 3.86)} }{2(4.9)} \\\\t = 2.14 \ s \ \ or \ \ 0.37 \ s[/tex]
The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.
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