A ball is dropped onto the floor from a height of 20m. It rebounds to a height of 15m. If the ball was in contact with the floor for 0.01sec what was its average acceleration during the contact take g=10m/s
To find the average acceleration of the ball during the contact with the floor, we can use the formula:
average acceleration = change in velocity / time taken
First, let's find the change in velocity. Since the ball falls from a height of 20m and rebounds to a height of 15m, the change in velocity equals the speed it takes to fall from 20m to 15m.
Using the equation for the velocity of a falling object:
v = √(2 * g * h)
where v is the velocity, g is the acceleration due to gravity (10m/s), and h is the height.
For the initial fall from 20m:
v_1 = √(2 * 10 * 20) = √(400) = 20m/s (downward)
For the rebound to 15m:
v_2 = √(2 * 10 * 15) = √(300) ≈ 17.32m/s (upward)
Now that we have the initial velocity (20m/s) and the final velocity (17.32m/s), we can find the change in velocity:
change in velocity = final velocity - initial velocity = 17.32m/s - (-20m/s) = 37.32m/s.
The time taken for the contact is given as 0.01s.
Now we can calculate the average acceleration:
average acceleration = change in velocity / time taken = 37.32m/s / 0.01s = 3732m/s^2.
Therefore, the average acceleration of the ball during the contact with the floor is 3732m/s^2.
downward velocity at impact
... v = √(2 * 10 * 20) = 20 m/s
upward velocity after impact
... v = √(2 * 10 * 15) = 10√3 m/s
accel = Δv / t = (20 + 10√3) / .01
...units are m/s²