A ball is dropped 2.75 m to the floor below. What is the acceleration of the ball while it is on the floor if it is on the floor for 0.0400 s?
I'm really confused on this question because it is a free fall question so wouldn't the acceleration be -9.81 m/s/s? How is the answer +340.0 m/s/s?
This is what I have but I don't know which equation to use...
y = -2.75 m
vi = 0 m/s
vf = ?
a = -9.81 m/s/s
t = 0.0400 s
Someone responded but I keep on getting the wrong answer. I'm getting 367.5 m/s/s and not +340.0 m/s/s.
Well, it seems like someone dropped the ball with this question. Let me help you clarify things. You are correct in assuming that the acceleration due to gravity is approximately -9.81 m/s². This negative sign indicates that the acceleration is directed downwards.
Now, let's think about what's happening when the ball hits the floor. The ball is initially in free fall, but upon hitting the floor, it experiences a sudden change in velocity. For a very brief moment, the ball comes to a momentary stop before rebounding back up, right?
During this instantaneous stop, the ball experiences a sudden change in velocity, which we can call Δv. Now, the time it takes for the ball to come to a stop and rebound can be regarded as a very short interval, let's call it Δt. During this interval, the ball's final velocity would be 0 m/s.
Since you have the initial velocity (vi = 0 m/s), final velocity (vf = 0 m/s), and time (Δt = 0.0400 s), the equation to calculate acceleration is:
a = Δv/Δt
Since the ball experiences a change in velocity from 0 m/s to 0 m/s, we have:
a = (0 - 0) m/s / 0.0400 s
Wait a minute! Division by zero? That's a big no-no in mathematics. Since we can't divide by zero, we can't calculate the acceleration during that momentary stop. So, the answer should be that the acceleration is undefined during that interval.
I hope this clears up the confusion and helps you avoid any further "divisions" in the future!