Question

A 0.28-kg stone you throw rises 34.3 m in the air. The magnitude of the impulse the stone received from your hand while being thrown is

Answer 1

Answer:

7.3 kg m/s

Explanation:

First of all, let's calculate the gravitational potential energy of the stone as it reaches its highest point:

$U=mgh=(0.28 kg)(9.8 m/s^2)(34.3 m)=94.1 J$

For the law of conservation of energy, this is equal to the initial kinetic energy of the stone at ground level (where the potential energy is zero), just after the stone leaves your hand:

$K=\frac{1}{2}mv^2=94.1 J$

From this equation we can find the velocity of the stone as it leaves your hand:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(94.1 J)}{0.28 kg}}=25.9 m/s$

The initial velocity of the stone (before leaving your hand) is zero:

$u=0$

The impulse received by the stone is equal to its change in momentum, so:

$I=\Delta p=m\Delta v=m(v-u)=(0.28 kg)(25.9 m/s-0)=7.3 kg m/s$