A student kicks a football vertically upwards, causing the football to reach a maximum height of 40.0m before falling back to the ground. Determine all unknowns and answer the following questions. Neglect drag and the initial height and horizontal motion of the football.A) With what speed was the football kicked?
B) What was the football's total flight time?
Answers
Answer:
A. 28 m/s
B. 5.72 s
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
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c.
scales
b.
overtones
d.
percussion
Answers
Answer:
overtones
Explanation:
Answers
Answer:
Explanation:
Your initial velocity () is 10 m/s, your time is 20 seconds. Although it is not explicitly mentioned in the problem, since you know the car comes to a stop, your final velocity (
) is 0 m/s.
The equation you can use to solve for acceleration (a) is
1. Plug in the numbers you do have, so your equation will look like this:
0=10+a(20)
2. Solve for a.
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Answers
Answer:
Force exerted on the car is 7030 N.
Explanation:
It is given that,
Mass of the car, m = 740 kg
Initial speed of the car, u = 19 m/s
Final speed of the car, v = 0
Time taken, t = 2 s
Let F is the force exerted on the car during this stop. We know that it is equal to the product of force and acceleration. Mathematically, it is given as :
F = -7030 N
So, the force exerted on the car during this stop is 7030 N. Hence, this is the required solution.
acceleration = change_in_velocity / time
so:
force = 740 kg * (19 m/s - 0 m/s) / 2.0 s
= 740 * 19 / 2 kg m per second^{2}
= 7030 kg m per second^{2}
= 7030 newtons of force