Question

Homer agin leads the varsity team in home runs. In a recent game, homer hit a 96 mi/hr sinking curve ball head on, sending it off his bat in the exact opposite direction at 56 mi/hr. The actual contact between ball and bat lasted for 0.75 milliseconds. Determine the magnitude of the average acceleration of the ball during the contact with the bat. Express your answer in m/s/s.

Answer 1

Answer;

= 90600 m/s...

Explanation;

velocity of ball = 96 mi/hr = 42.916 m/s...  

velocity of bat = -56mi/hr = -25.034m/s.. ( note the negativity here..)  

time = 0.75 millisec = 0.75 * 10^-3 sec  

acceleration = (velball - velbat) / time

                     = (42.916 - (-25.034)) / (0.75 * 10^-3)

                     =90600 m/s

Answer 2

Answer:

The magnitude of the average acceleration of the ball during the contact with the bat is 90666m/s^2

Explanation:

Average acceleration (a) is defined as the change in velocity divided by the time it took:

$a=\frac{V_{2}-V_{1} }{t}$ (eq. 1)

V2: final velocity

V1: initial velocity

t: time the change in velocity took

In this example, the ball traveled at 96mi/h in a direction and changed its velocity to 56mi/h in the opposite direction because of the contact with the bat.

In this case, we need to express all in meters and seconds. To achieve this use some conversion factors:

$V_{1} =96mi/h=96mi/h*(\frac{0.45m/s}{1mi/h} )=43m/s\\V_{2} =-56mi/h=-56mi/h*(\frac{0.45m/s}{1mi/h} )=-25m/s\\t=0.75ms=0.75ms*(\frac{1s}{1000ms} )=7.5*10^{-4} s$

Note there's a negative sign beside 56 mi/hr because it goes in the opposite direction to 96 mi/hr.

Plugging in these values in eq. 1:

$a=\frac{-25m/s-43m/s }{7.5*10^{-4} s}=90666m/s^2$