A 0.15 kg baseball moving at +26 m/s is slowed to a stop by a catcher who exerts a constant force of -390 N. How long does it take this force to stop the ball?

0.01 s

Explanation:

The following data were obtained from the question:

Mass (m) of baseball = 0.15 kg

Initial velocity (u) = +26 m/s

Final velocity (v) = 0 m/s

Force (F) applied = –390 N

Time (t) =?

Next, we shall determine the acceleration of the baseball. This can be obtained as follow:

Mass (m) of baseball = 0.15 kg

Force (F) applied = –390 N

Acceleration (a) of baseball =?

F = ma

–390 = 0.15 × a

Divide both side by 0.15

a = –390/0.15

a = –2600 m/s²

Finally, we shall determine the time taken for the baseball to stop. This can be obtained as follow:

Initial velocity (u) = +26 m/s

Final velocity (v) = 0 m/s

Acceleration (a) = –2600 m/s²

Time (t) =?

v = u + at

0 = 26 + (–2600 × t)

0 = 26 – 2600t

Collect like terms

0 – 26 = – 2600t

– 26 = – 2600t

Divide both side by – 2600

t = –26 / –2600

t = 0.01 s

Thus, it will take 0.01 s for the baseball to stop.

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr