Question

a baseball travelling at 26 M S is caught by catcher if it travels 9.0 cm into the mitt and comes to a stop what is the deceleration of the ball

Answer 1

We are given the following information

Initial velocity of ball = 26 m/s

Distance traveled by ball = 9 cm = 9/100 = 0.09 m

We are asked to find the deceleration of the ball.

Recall the third equation of the motion given by

`v^2_f=v^2_i+2as`

Where a is the deceleration, s is the distance traveled, vi is the initial velocity and vf is the final velocity of the ball.

vf = 0 since the ball is stopped after getting caught by the catcher.

Let us substitute the given values into the above equation

`\begin{gathered} v^2_f=v^2_i+2as \\ 0=26^2_{}+2\cdot a\cdot0.09 \\ 0=676+0.18\cdot a \\ 0.18\cdot a=-676 \\ a=-(676)/(0.18) \\ a=-3755.56\; (m)/(s^2) \end{gathered}`

Therefore, the deceleration of the ball is -3755.56 m/s^2

The negative sign indicates deceleration.