a golf ball released from a height of 1.5 m above aa concreate floor bounces back to a height of 1.1m if the ball is in contact with the floor for .62 ms what is the average acceleartion of the ball while in contact with the floor?
Vf^2=2ad
so in each of these, 1.5 meter, 1.1m find the floor velocity involved. Notice that they are opposite directions. a is g.
Now, having those two velocities, V1, V2
a=(V2-V1)/t but working in just speeds, and knowing they are in opposite directions
a=(V2+V1)/t
To find the average acceleration of the ball while in contact with the floor, we need to use the equation for average acceleration:
average acceleration = change in velocity / time
In this case, the ball is initially at rest, so the change in velocity is given by the final velocity. Since the ball bounces back to a height of 1.1m, we can use the relationship between velocity and height in an idealized bouncing scenario:
final velocity = √(2 * g * height)
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and height is the maximum height the ball reaches (1.1m in this case).
Using this formula, we can find the final velocity:
final velocity = √(2 * 9.8 * 1.1)
Next, we need to calculate the time of contact with the floor. We are given that the ball is in contact with the floor for 0.62 ms, which we need to convert to seconds:
time = 0.62 ms = 0.62 * 10^(-3) s
Now, we can calculate the average acceleration:
average acceleration = final velocity / time
Plug in the values to get the answer.