Question

1. a well - thrown ball is caught in a well - padded mitt. if the deceleration of the ball is 2.10×10^1 m/s², and 1.85 ms (1 ms = 10^(-3) s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v = v_0+at$, where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. Since the ball stops, $v = 0$. The acceleration $a=- 2.10\times10^{1}\ m/s^{2}$ (negative because it is deceleration) and $t = 1.85\times10^{-3}\ s$.

Step2: Rearrange the kinematic - equation for $v_0$

From $v = v_0+at$, we can solve for $v_0$: $v_0=v - at$. Substituting $v = 0$, $a=-2.10\times 10^{1}\ m/s^{2}$ and $t = 1.85\times10^{-3}\ s$ into the equation.

Step3: Calculate $v_0$

$v_0=0-(-2.10\times 10^{1}\ m/s^{2})\times(1.85\times10^{-3}\ s)$.
$v_0=(2.10\times 10^{1}\ m/s^{2})\times(1.85\times10^{-3}\ s)$.
$v_0 = 21\times1.85\times10^{-2}\ m/s$.
$v_0=0.3885\ m/s$.

Answer:

$0.3885\ m/s$