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A baseball ball, of 0.22 kg, is launched with a velocity of 32 m/s horizontally. A batter hits the ball, and this take a velocity of 61 m/s at 40 degrees above the horizontal. Which is the momentum applied to the ball?

**Solution with FisicaLab**

Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, one element Mobile, and one element Momentum. As show the image below:

We don’t have a time data to the impact (the duration of the impact), but we assume 0.01 seconds, a reasonable time. This time is necessary to calculate the impulsive force in the element Momentum. Then, to the element Stationary reference system, write:

**g**9.81

**t**0.01

We assume that the ball is, initially, moving in the direction of negative X. Then in the element Mobile, we call this *Ball*, write:

**Name**Ball

**m**0.22

**vi**32

**angi**180

**vf**61

**angf**40

To the element Momentum:

**Object**Ball

**Imp**imp

**ang**ang

**fImp**f

Now click in the icon Solve to get the answer:

imp = 9.215 N*s ; ang = 69.412 degrees ; f = 921.472 N ; Status = success.