at a ball game consider a baseball of mass m=0.15 that falls directly downward at a speed v=40m/s into the hands of a fan. what impulseft must be supplied to bring the ball to rest? if the ball is stopped in 0.3s, what is the average force of the ball on the catchers hand?

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To find the impulse required to bring the ball to rest, we can use the formula for impulse:

Impulse = change in momentum

The formula for momentum is given by:

Momentum = mass × velocity

Since the ball is falling directly downward, we consider the direction of motion to be positive.

Given:
Mass of the baseball, m = 0.15 kg
Initial velocity of the baseball, v = 40 m/s (downward)

First, let's find the momentum of the ball before it was caught. The initial momentum is given by:

Initial momentum = mass × initial velocity

Substituting the values:

Initial momentum = 0.15 kg × (-40 m/s) [Negative sign due to downward motion]

Next, we know that the ball comes to rest, so its final momentum is zero. Thus, the change in momentum is:

Change in momentum = Final momentum - Initial momentum
= 0 - (0.15 kg × (-40 m/s))
= 6 kg·m/s

Therefore, the impulse required to bring the ball to rest is 6 kg·m/s.

Now, let's calculate the average force exerted on the catcher's hand using the equation:

Impulse = Force × time

Given:
Impulse = 6 kg·m/s
Time taken, t = 0.3 s

Rearranging the equation, we have:

Force = Impulse / time

Substituting the values:

Force = 6 kg·m/s / 0.3 s
= 20 N

Hence, the average force of the ball on the catcher's hand is 20 Newtons.