Share
A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does the ball have
Question
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Answers ( )
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
(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
Time of flight, T = 2t
(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