Is 0.2 kg softball is thrown towards a catches mid the ball is accelerating at a rate of 8 m/s squared what was for the ball hit the catches

Answer 1

To find the time it takes for the ball to hit the catcher, we can use the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

Where:
s = displacement (which is 0.2 kg since the ball is caught)
u = initial velocity (0 m/s since the ball starts from rest)
a = acceleration (8 m/s^2)
t = time

Plugging in the values, we have:

\(0.2 = 0 \cdot t + \frac{1}{2} \cdot 8 \cdot t^2\)

Simplifying, we get:

\(0.2 = 4t^2\)

Divide both sides by 4:

\(0.05 = t^2\)

Taking the square root of both sides:

\(t = \sqrt{0.05}\)

So, the time it takes for the ball to hit the catcher is approximately 0.224 seconds.