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Makovka662 [10]
3 years ago
14

Safawi kicked a ball with an initial speed of 20 ms) at 30° angle. If the ball experienced a constant vertical acceleration of -

9.81 ms?, calculate the ball's maximum height and distance. Explain how soccer player could apply the force of summation theory to optimize his kicks performance.
(10 marks)​
Physics
1 answer:
Alex73 [517]3 years ago
3 0

Answer:

maximum height = 5.1 m  distance traveled/horizontal displacement = 35.3m

Explanation:

Vy = sin 30º • 20 m/s

Vy = 10 m/s

Vx = cos 30º • 20m/s

Vx = 17.3

V final = V initial + acceleration • time

0 = 10m/s + (-9.81 m/s/s) • time

subtract 10 m/s from both sides

-10m/s = -9.81m/s/s • time

divide each side by -9.81m/s/s

t = 10/9.81 s or roughly 1.02 second  

the time to reach max height is equal to 1.02 s

Height = 1/2 • acceleration • time^2  + Vy • time

Height = 1/2 • (-9.81 m/s/s) •  1.02s^2  + 10m/s • 1.02 s

Height = 5.1 m

the maximum height of the ball is equal to 5.1m

since the ball starts from an initial height of zero and has a final height of 0 once it reaches the ground, the time needed to reach max height is the same amount of time to fall back down to earth. thus, time total is equal to 1.02s multiplied by 2

t total = 1.02s • 2

t total = 2.04s

horizontal displacement = Vx • time total

horizontal displacement = 17.3m/s • 2.04 s

Horizontal displacement = 35.3m

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a substance dissolves.  


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usually adding a mixture to a mixture has little energy change, i.e little heat taken in by the reaction mixture or little heat given out by the reaction mixture. Whereas when a new substance is formed, there is usually noticeable energy change like the container gets colder or hotter (without heat being supplied of course). For example dissolving basic oxides into water releases energy ( more energy released than gained = exothermic reaction).  


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The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss} (1)

From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

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W_{loss} = 6574.75\,J

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