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
A) 0.27 N s.
B) 2.7 N s.
C) 7.3 N s.
D) 9.6 N s.
E) 34.3 N s.

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
A) 0.27 N s. 
B) 2.7 N s. 
C) 7.3 N s. 
D) 9.6 N s. 
E) 34.3 N s.

Quizplus · Accepted Answer

We can use the formula for impulse:

$J = \Delta p = m\Delta v$

where $J$ is the magnitude of the impulse, $\Delta p$ is the change in momentum, $m$ is the mass of the stone, and $\Delta v$ is the change in velocity.

At the highest point of its trajectory, the stone's velocity is zero. Therefore, the change in velocity is:

$\Delta v = vf - vi = 0 - vf = -vf$

where $vf$ is the final velocity (zero) and $vi$ is the initial velocity.

Using the kinematic equation:

$v_f^2 - v_i^2 = 2ad$

where $a$ is the acceleration (due to gravity, approximately -9.8 m/s^2) and $d$ is the displacement (the height the stone rises, 34.3 m), we can solve for $v_i$:

$v_i = \sqrt{v_f^2 - 2ad} = \sqrt{0 - 2(-9.8)(34.3)} \approx 23.0 	ext{ m/s}$

Therefore, the change in velocity is:

$\Delta v = -vf = -23.0 	ext{ m/s}$

Finally, we can calculate the impulse:

$J = m\Delta v = (0.28 	ext{ kg})(-23.0 	ext{ m/s}) \approx -6.44 	ext{ N s}$

The magnitude of the impulse is 6.44 N s, but since the question asks for the magnitude of the impulse (and not the signed value), the answer is:

$J \approx 6.44 	ext{ N s} \approx \boxed{	extbf{(C) }7.3 	ext{ N}}$

A 028-Kg Stone You Throw Rises 34