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
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.