Answer:
1. 53.415 m
2. 32.373 m/s
3. 30.82 m/s
Explanation:
Let g = 9.81 m/s2. We can use the following equation of motion to calculate the distance traveled by the ball in 3.3s, and the velocity it achieved
1.[tex]s = gt^2/2 = 9.81*3.3^2/2 = 53.415 m[/tex]
2.[tex]v = gt = 9.81*3.3 = 32.373 m/s[/tex]
3. If the ball is kicked at 0.5 m above the ground then the net distance between the ball and the roof top is
53.415 - 0.5 = 48.415 m
For the ball to at least make it to the roof top at speed v = 0 m/s. We can use the following equation of motion to calculate the minimum initial speed
[tex]v^2 - v_0^2 = 2g\Delta s[/tex]
where v = 0 m/s is the final velocity of the ball when it reaches the rooftop, [tex]v_0[/tex] is the initial velocity, [tex]\Delta s = 48.415[/tex] is the distance traveled, g = -9.81 is the gravitational acceleration with direction opposite with velocity
[tex]0 - v_0^2 = 2*(-9.81)*48.415[/tex]
[tex]v_0^2 = 950[/tex]
[tex]v_0 = \sqrt{950} = 30.82 m/s[/tex]