To solve this problem, we need to find the force of the impact when the ball is caught, using the data provided. We'll use the following formula:
[tex]\[ F = ma \][/tex]
where [tex]\( F \)[/tex] is the force, [tex]\( m \)[/tex] is the mass of the object, and [tex]\( a \)[/tex] is the acceleration.
We already know the mass [tex]\( m \)[/tex] of the softball is 0.2 kilograms.
Next, we need to find the acceleration [tex]\( a \)[/tex].
Acceleration can be calculated using the formula:
[tex]\[ a = \frac{{v - u}}{t} \][/tex]
where:
- [tex]\( v \)[/tex] is the final velocity,
- [tex]\( u \)[/tex] is the initial velocity,
- [tex]\( t \)[/tex] is the time.
Given:
- The initial velocity [tex]\( u \)[/tex] is 32 meters/second,
- The final velocity [tex]\( v \)[/tex] is 0 meters/second,
- The time [tex]\( t \)[/tex] taken to stop is 0.8 seconds.
Let's calculate the acceleration [tex]\( a \)[/tex] first:
[tex]\[ a = \frac{{v - u}}{t} = \frac{0 - 32}{0.8} = \frac{-32}{0.8} = -40 \text{ meters per second squared} \][/tex]
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, meaning it's a deceleration.
Now, we can calculate the force [tex]\( F \)[/tex]:
[tex]\[ F = ma = 0.2 \text{ kg} \times (-40) \text{ m/s}^2 = -8 \text{ newtons} \][/tex]
The force of the impact, taking into account the direction, is -8 newtons. However, when reporting the magnitude of the force, we typically ignore the negative sign (it indicates direction, which is often implied in the context of deceleration).
Therefore, the magnitude of the force of the impact is:
[tex]\[ \boxed{8} \][/tex]
So, the force of the impact is 8 newtons.
