8. As you throw a 2.00 x 102 g baseball, your hand moves a distance of 80.0 cm before you release the ball. If you exert an average force of 55.0 N on the ball while it is in your hand, determine:

Respuesta :

Answer:

v = 21.0 m / s

Explanation:

In this exercise, it is generally asked to determine the final speed of the ball.

Let's start by finding the acceleration of it

            F = ma

            a = F / m

let's reduce the magnitudes to the SI system

            m = 2.00 102 g (1kg / 1000g) = 0.200 kg

            x = 80.0 cm (1m / 100 cm) = 0.800 m

we calculate

            a = 55.0 / 0.200

            a = 275 m / s

now we can use kinematics, where the ball has an initial velocity of zero and travels a distance x = 80 cm with the force

            v² = v₀² + 2 a x

            v² = 0 + 2ax

            v = [tex]\sqrt{2ax}[/tex]

let's calculate

           v = [tex]\sqrt{2 \ 275 \ 0.800}[/tex]Ra 2 275 0.800

           v = 20.98 m / s

if we use the criteria of significant figures

           v = 21.0 m / s