ball is thrown upward with a velocity of $19.6 \mathrm{~m} / \mathrm{s}$. What is its velocity after $3.00 \mathrm{~s}$ ?
A) $9.80 \mathrm{~m} / \mathrm{s}$ down
B) 19.6 down
C) zero
D) $9.80 \mathrm{~m} / \mathrm{s}$ up

Question

Quizplus · Accepted Answer

The velocity of the ball after 3.00 s can be calculated using the equation $v = u + at$, where $v$ is the final velocity, $u$ is the initial velocity (19.6 m/s upwards), $a$ is the acceleration due to gravity (-9.8 m/s$^2$ downwards), and $t$ is the time (3.00 s). Substituting the values, we get $v = 19.6 m/s + (-9.8 m/s^2)(3.00 s) = 19.6 m/s - 29.4 m/s = -9.8 m/s$, indicating a velocity of $9.80 m/s$ downwards.

Ball Is Thrown Upward with a Velocity Of 19.6 m/s19.6 \mathrm{~m} / \mathrm{s}19.6 m/s

Ball Is Thrown Upward with a Velocity Of $19.6 \mathrm{~m} / \mathrm{s}$