Answer:
The value is [tex]\theta = 34.8 ^o[/tex]
Explanation:
Generally the time it takes the ball to move from maximum height to the ground is
[tex]t = \frac{v - u}{g}[/tex]
Here u= 0 m/s and v = 15 m/s
So
[tex]t = \frac{15 - 0}{9.8}[/tex]
=> [tex]t = 1.53 \ s [/tex]
Generally the maximum height traveled in time t is mathematically represented as
[tex]s = u * t + \frac{1}{2} * g * t^2[/tex]
=> [tex]s = 0 * 1.53 + \frac{1}{2} * 9.8 * 1.53^2[/tex]
=> [tex]s = 11.5 \ m [/tex]
The interval in distance between the maximum height and the initial position of the ball is
[tex]d = s - k[/tex]
Here k = 5 m (given)
[tex]d = 11.5 - 5[/tex]
[tex]d = 6.5 \ m [/tex]
Generally from the kinematic equation we have that
[tex]d = ut_1 + \frac{1}{2} g * t_1 ^2[/tex]
Here [tex] t_1 [/tex] is the time taken to travel through d
[tex]t_1 = \sqrt{ \frac{2 * d }{g} }[/tex]
=> [tex]t_1 = \sqrt{ \frac{2 * 6.5 }{9.8} }[/tex]
=> [tex]t_1 = 1.15 \ s [/tex]
Generally the total time for this fight is mathematically represented as
[tex]t_f = t + t_1[/tex]
=> [tex]t_f = 1.15 + 1.53[/tex]
=> [tex]t_f = 2.68 \ s [/tex]
Generally the vertical force exerted on the cube is mathematically represented as
[tex]F = m * g * sin (\theta )[/tex]
=> [tex] m * a = m * g * sin (\theta )[/tex]
=> [tex] a = g * sin (\theta )[/tex]
Now a is mathematically represented as
[tex]a = \frac{v - u }{t_f}[/tex]
=> [tex]a = \frac{15 - u }{2.68}[/tex]
=> [tex]a = 5.597 \ m/s^2[/tex]
So
[tex] 5.597 = 9.8 * sin (\theta )[/tex]
[tex]\theta = sin^{-1} [\frac{5.597}{9.8} ][/tex]
[tex]\theta = 34.8 ^o[/tex]
