a baseball is thrown at 30ms^-1. the batter hits it and gives it a velcoity of 40ms^-1 in the opposite direction. if the ball and bat are in contact for 0.04 s, what is the average acceleration of the ball during this interval?
To find the average acceleration of the ball during the interval it is in contact with the bat, you need to know the change in velocity and the time taken.
Given:
Initial velocity of the ball (u) = 30 m/s (thrown)
Final velocity of the ball (v) = -40 m/s (opposite direction after being hit)
Time taken (t) = 0.04 seconds
The change in velocity (Δv) is the difference between the final velocity and the initial velocity.
Δv = v - u
Δv = (-40 m/s) - (30 m/s)
Δv = -70 m/s
The average acceleration (a) can be calculated using the formula:
a = Δv / t
Substituting the values, we get:
a = (-70 m/s) / (0.04 s)
a ≈ -1750 m/s²
Therefore, the average acceleration of the ball during the interval it is in contact with the bat is approximately -1750 m/s².
V = Vo + a*t
V = -40 m/s
Vo = 30 m/s
t = 0.04 s.
Solve for a(It is negative).