A 0.330-kg volleyball is thrown vertically downward with a speed of 0.150 m/s in a place where g = $9.81 \mathrm {~m} / \mathrm { s } ^ { 2 }$ It takes it 0.0655 s to reach the ground. What is the magnitude of its momentum just before it hits the ground?
A) $0.0418 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
B) $0.0216 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $0.212 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $0.163 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $0.262 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Question

A  0.330-kg volleyball is thrown vertically downward with a speed of  0.150 m/s in a place where  g = $9.81 \mathrm {~m} / \mathrm { s } ^ { 2 }$ It takes it  0.0655 s  to reach the ground. What is the magnitude of its momentum just before it hits the ground?
A) $0.0418 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
B) $0.0216 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $0.212 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $0.163 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $0.262 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Quizplus · Accepted Answer

To find the momentum just before the volleyball hits the ground, we first calculate its final velocity using the equation $v = u + at$, where $v$ is the final velocity, $u$ is the initial velocity (0.150 m/s), $a$ is the acceleration due to gravity (9.81 m/s^2), and $t$ is the time (0.0655 s). Plugging in the values, we get $v = 0.150 m/s + (9.81 m/s^2)(0.0655 s) = 0.792 m/s$. The momentum ($p$) is given by $p = mv$, where $m$ is the mass (0.330 kg) and $v$ is the final velocity. Thus, $p = 0.330 kg \cdot 0.792 m/s = 0.26136 kg \cdot m/s$, which rounds to $0.262 kg \cdot m/s$, making option E correct.

A 0330-Kg Volleyball Is Thrown Vertically Downward with a Speed of Speed