Answer:
A. 28 m/s
B. 5.72 s
Step-by-step explanation:
The following data were obtained from the question:
Maximum height (h) = 40 m
Acceleration due to gravity (g) = 9.8 m/s²
A. Determination of the initial velocity.
Maximum height (h) = 40 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity at maximum height (v) = 0 m/s
Initial velocity (u) =.?
v² = u² – 2gh (Ball is going against gravity)
0² = u² – (2 × 9.8 × 40)
0 = u² – 784
Collect like terms
0 + 784 = u²
784 = u²
Take the square root of both side
u = √784
u = 28 m/s
Thus, the ball was kicked with a velocity of 28 m/s
B. Determination of the time of flight.
We'll begin by calculating the time taken to reach the maximum height. This can be obtained as follow:
Velocity at maximum height (v) = 0 m/s
Initial velocity (u) = 28 m/s
Time taken to reach the maximum height (t) =?
v = u – gt (Ball is going against gravity)
0 = 28 – 9.8t
Rearrange
0 + 9.8t = 28
9.8t = 28
Divide both side by 9.8
t = 28/9.8
t = 2.86 s
Finally, we shall determine the the time of flight as follow:
Time taken to reach the maximum height (t) = 2.86 s
Time of flight (T) =?
T = 2t
T = 2 × 2.86
T = 5.72 s
Therefore, the time of flight for the ball is 5.72 s