Answer:
(A) The time taken for the ball to stop is 8.7 x 10⁻³ s
(B) The distance traveled by the baseball before stopping is 0.3 m
Explanation:
Given;
mass of the baseball, m = 0.14 kg
velocity of the baseball, v = 23 m/s
force exerted on the baseball by the catcher, F = 370 N
(A) The time taken for the ball to stop;
[tex]F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.14 \times 23}{370} \\\\t = 8.7\times 10^{-3} \ s\\\\t = 8.7 \ ms[/tex]
(B) The distance traveled by the baseball before stopping is calculated as;
acceleration of the ball, [tex]a = \frac{v}{t} = \frac{23}{8.7\times 10^{-3}} = 2643.678 \ m/s^2[/tex]
Distance traveled, s;
s = ut + ¹/₂at²
s = (23)(8.7 x 10⁻³) + ¹/₂(2643.678)(8.7 x 10⁻³)²
s = 0.2001 + 0.1001
s = 0.3 m
