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erik [133]
3 years ago
11

If a projectile was launched at an angle of 13 degrees above the ground and hit its target 14.3 m away, what was the projectile'

s launch speed? Ignore air resistance and acceleration due to gravity is 9.8m/s^2.
Physics
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

18 m/s

Explanation:

Range of a projectile on level ground is:

R = v₀² sin(2θ) / g

14.3 m = v₀² sin(2×13°) / 9.8 m/s²

v₀ = 17.9 m/s

Rounded to two significant figures, the launch speed was 18 m/s.

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Answer:

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Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

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With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

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V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}

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