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A 2.50-kg stone is dropped from rest at a height of 3.75 m. what impulse does gravity impart to this stone from the instant it is dropped until it hits the ground, assuming negligible air resistance?

Verified Answer :

Answer:

The impulse is 21.41 kg-m/s.

Explanation:

Given that,

Mass of stone = 2.50 kg

Height = 3.75 m

We need to calculate the time

Using equation of motion

\(s=ut+\dfrac{1}{2}gt^2\)

Where, s = height

u = initial speed

t = time

Put the value in the equation

\(3.75=0+\dfrac{1}{2}\times9.8\times t^2\)

\(t^2=\dfrac{3.75\times2}{9.8}\)

\(t=\sqrt{\dfrac{3.75\times2}{9.8}}\)

\(t=0.874\ sec\)

We need to calculate the impulse

Using formula of impulse

\(J=F\times t\)

\(J=mg\times t\)

Put the value into the formula

\(J=2.50\times9.8\times0.874\)

\(J=21.41\ kg-m/s\)

Hence, The impulse is 21.41 kg-m/s.

The impulse gravity impart to this stone from the instant it is dropped until it hits the ground is 21.41 kg-m/s.

What is an impulse?

The impulse is defined as the effect of the force over a time. Like if you fall in a hard surface then force will be higher as compared to if you fall on a soft net. In soft net, the time increases and the effect of force decreases.

It is given that,

Mass of stone = 2.50 kg

Height = 3.75 m

We need to calculate the time

Using equation of motion

\(s=ut + \frac{1}{2} at^2\)

Where, s = height

u = initial speed

t = time

Put the value in the equation

\(3.75=0+\frac{1}{2} \times 9.8\times t^2\)

\(t=\sqrt{\dfrac{3.75\times 2}{9.8}\)

\(t=0.874\ sec\)

We need to calculate the impulse

Using formula of impulse

\(J=F\times t\)

\(J=mg\times t\)

Put the value into the formula

\(J=2.5\times 9.8\times 0.874\)

\(J=21.41 \ kg-\frac{m}{s}\)

Hence,the impulse gravity impart to this stone from the instant it is dropped until it hits the ground is 21.41 kg-m/s.

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