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Stells [14]
2 years ago
12

2. FAt=p (remember p=mv). A 2.5Jg baseball moving at 8.5m/s to the left is slowed by a tipped

Physics
1 answer:
TiliK225 [7]2 years ago
3 0

The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 8.5 m/s
  • Final velocity (v) = 7.5 m/s
  • Time (t) = 5 ms = 0.25 s
  • Mass (m) = 2.5 Kg
  • Force (F) = ?

<h3>How to determine the force</h3>

The force exerted on the ball can be obtained as follow:

F = m(v + u) / t

F = [2.5(7.5 + 8.5)]/ 0.25

F = 40 / 0.25

F = 160 N

Thus, the force exerted on the ball is 160 N

Learn more about momentum:

brainly.com/question/250648

#SPJ1

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