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A softball pitcher swings a ball of mass 0.244 kg around a vertical circular path of radius 59.3 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.3 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 14.7 m/s. If the ball is released at the bottom of the circle, what is its speed upon release

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nuhulawal20

Answer: its speed upon release is 26.05 m/s

Explanation:

Given that;

mass m = 0.244 kg

force F = 30.3 N

V1 = 14.7 m/s

r = 59.3 cm = 0.593 m

Vf = ?

we know that;

1/2mV1² + FπR = 1/2mVf²

so we substitute

[1/2×0.244×(14.7)²] + [30.3×π×0.593 = 1/2×0.244×Vf²

26.3629 + 56.4478 = 0.122Vf²

82.8107 = 0.122Vf²

Vf² = 82.8107 / 0.122Vf

Vf² = 678.7762

Vf = √678.7762

Vf = 26.05 m/s

Therefore its speed upon release is 26.05 m/s

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