Answer:
(A) The maximum height of the ball is 40.57 m
(B) Time spent by the ball on air is 5.76 s
(C) at 33.23 m the speed will be 12 m/s
Explanation:
Given;
initial velocity of the ball, u = 28.2 m/s
(A) The maximum height
At maximum height, the final velocity, v = 0
v² = u² -2gh
u² = 2gh
[tex]h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m[/tex]
(B) Time spent by the ball on air
Time of flight = Time to reach maximum height + time to hit ground.
Time to reach maximum height = time to hit ground.
Time to reach maximum height is given by;
v = u - gt
u = gt
[tex]t = \frac{u}{g}[/tex]
Time of flight, T = 2t
[tex]T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s[/tex]
(C) the position of the ball at 12 m/s
As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.
v² = u² - 2gh
12² = 28.2² - 2(9.8)h
12² - 28.2² = - 2(9.8)h
-651.24 = -19.6h
h = 651.24 / 19.6
h = 33.23 m
Thus, at 33.23 m the speed will be 12 m/s
