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A ball of mass m is dropped. What is the formula for the impulse exerted on the ball from the instant it is dropped to an arbitrary time τ later? Ignore air resistance.

a) ( {Impulse} = m cdot tau )
b) ( {Impulse} = m cdot g cdot tau )
c) ( {Impulse} = m cdot Delta v )
d) ( {Impulse} = F cdot tau )

User Gedeon
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Final answer:

The impulse exerted on a dropped ball of mass m over time τ is calculated by the formula (Impulse = m · g · τ), assuming the only force acting is gravity.

Step-by-step explanation:

The formula for the impulse exerted on a ball of mass m from the instant it is dropped to an arbitrary time τ later, ignoring air resistance, is (Impulse = m · g · τ). This formula assumes that the only force acting on the ball while it is falling is gravity, which accelerates the ball at a constant rate g.

The impulse is the product of the average force over the time interval it acts, and since the only force here is gravity, the impulse can be calculated by multiplying the force due to gravity (m · g) by the time τ.

User Delixfe
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