A 0.330-kg volleyball is thrown vertically downward with a speed of 0.150 m/s in a place where g = 9.81 m/s².It takes it 0.0655 s to reach the ground.What is the magnitude of its momentum just before it hits the ground? A)0.212 kg ∙ m/s B)0.262 kg ∙ m/s C)0.163 kg ∙ m/s D)0.0216 kg ∙ m/s E)0.0418 kg ∙ m/s

Question

Quizplus · Accepted Answer

To find the momentum just before the volleyball hits the ground, we first need to 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 downward), $a$ is the acceleration due to gravity (9.81 m/s$^2$ downward), and $t$ is the time (0.0655 s). Substituting the given values, we get $v = 0.150 m/s + (9.81 m/s$^2$) \times 0.0655 s = 0.792 m/s$. The magnitude of the momentum just before it hits the ground is given by $p = mv$, where $m$ is the mass of the volleyball (0.330 kg) and $v$ is the final velocity we just calculated. Therefore, $p = 0.330 kg \times 0.792 m/s = 0.26136 kg \cdot m/s$, which rounds to $0.262 kg \cdot m/s$, making option B the correct answer.

A 0330-Kg Volleyball Is Thrown Vertically Downward with a Speed of of 0.150