# Can someone explain this question using a FBD?

## Mar 25, 2016

Acceleration $a = - 5404.593 m {s}^{-} 2$ rounded to three decimal places. $- v e$ sign shows that it is retardation.

Force $f = 783.666 N$ rounded to three decimal places.

#### Explanation:

To simplify the problem we need to make the following assumptions: (Presume if these are understood, FBD may not be needed)

1. During the process of bringing the baseball to rest by Brandon, acceleration due to gravity $g$ plays negligible role. Hence it has been ignored.
2. The direction of motion of the ball and the direction in which force is applied by Brandon along the same line of action and therefore, full force is utilized in stopping the ball.
3. It is assumed that distance of recoil of hand is the distance moved by the baseball-hand system after the ball is captured.
4. This is the distance through which the ball comes to stop from initial velocity of $38.2 m {s}^{-} 1$ the moment the hand touched it first.
5. The force applied by the hand of player is such that constant retardation is produced during this distance.

(a) We know that that for constant acceleration $a$

${v}^{2} - {u}^{2} = 2 a s$
Where various symbols have meaning well known to us.
Inserting given values

${0}^{2} - {\left(38.2\right)}^{2} = 2 \times a \times 0.135$, solving for $a$
$a = - 5404.593 m {s}^{-} 2$ rounded to three decimal places. Negative sign shows that direction of acceleration is opposite to the direction of motion.
(b) Force $f = m \times a$
$\therefore f = 0.145 \times 5404.593 = 783.666 N$ rounded to three decimal places.