A baseball is thrown at 40 m/s. Jose Bautista hit it back in the opposite direction at a speed of 65 m/s. The ball is in contact for 1.2 ms. What distance does the ball travel while it is in contact with the bat?
a = change in velocity/time
= 105/1.2*10^-3 = (105/1.2)10^3
x = 40(1.2)10^-3 - (1/2)at^2
= 40*1.2*10^-3 -(105/2.4)10^3 (1.2)^2 10^-6
To find the distance the ball travels while in contact with the bat, we can use the formula:
Distance = Velocity × Time
The time in this case is given as 1.2 ms, which we need to convert to seconds.
1.2 ms = 1.2 × 10^(-3) s
Now, we can substitute the given values into the formula:
Distance = (Velocity of the ball + Velocity of Jose Bautista) × Time
The velocity of the ball is given as 40 m/s and the velocity of Jose Bautista is given as 65 m/s. We add these two velocities together:
Velocity = 65 m/s + (-40 m/s) [Since the ball is hit back in the opposite direction, we use a negative sign for the velocity of the ball.]
Velocity = 25 m/s
Now we have all the values to find the distance:
Distance = (25 m/s) × (1.2 × 10^(-3) s)
Calculating this, we get:
Distance = 0.03 m
So, the ball travels a distance of 0.03 meters while it is in contact with the bat.