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Makovka662 [10]
2 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]2 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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B. The value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

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Initial temperature T_{1} = 25 degree Celsius = 298 Kelvin

Final pressure P_{2} = 300 k pa

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⇒ V_{1} = V_{2} ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

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Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m C_{v} ( T_{2} -T_{1} )

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C_{v} = specific heat at constant volume of nitrogen = 0.743 \frac{KJ}{kg k}

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 \frac{KJ}{kg}

Therefore the value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

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