A golf ball released from a height of 1.80 m above a concrete floor, bounces back to a height of 1.06 m. If the ball is in contact with the floor for 4.62 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor?

Quidditch told me that the first steps would be to find the velocity as the ball first makes contact with the floor, and the velocity as the ball leaves the floor. But I'm still unsure how to do this...

THANK YOU!

To find the velocity as the ball first makes contact with the floor, we can use the equation:

v = u + at

Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time taken

In this case, the initial velocity u is 0 (as the ball is released from rest). We are trying to find the final velocity v, and the time taken t is given as 4.62 ms.

So, the equation becomes:

v = 0 + a * 0.00462

Simplifying, we get:

v = 0.00462a

Now, let's find the velocity as the ball leaves the floor. We can use the same equation:

v = u + at

In this case, the final velocity v is 0 (as the ball momentarily stops at the maximum height), the initial velocity u is the velocity at which the ball leaves the floor, and the time taken t is still 4.62 ms.

So, the equation becomes:

0 = u + a * 0.00462

Simplifying, we get:

u = -0.00462a

From the given information, the initial height (h1) is 1.80 m, and the final height (h2) is 1.06 m. The ball reaches its maximum height when its velocity becomes zero, so we can use the equation:

v^2 = u^2 + 2as

Where:
- v is the final velocity (0 m/s)
- u is the initial velocity
- a is the acceleration (which we need to find)
- s is the displacement (h2 - h1)

Plugging in the values, we get:

0^2 = u^2 + 2a(h2 - h1)

Simplifying, we get:

0 = u^2 + 2a(h2 - h1)

Substituting the previously found value for u (-0.00462a), we can write:

0 = (-0.00462a)^2 + 2a(h2 - h1)

Expanding and simplifying the equation, we have:

0 = 0.0000213444a^2 + 2a(1.06 - 1.80)

0 = 0.0000213444a^2 + 2a(-0.74)

0 = 0.0000213444a^2 - 1.48a

Now, we can factor out 'a' from the equation:

a(0.0000213444a - 1.48) = 0

Now, there are two possibilities for the average acceleration: either a = 0 or (0.0000213444a - 1.48) = 0.

If a = 0, then the magnitude of the average acceleration is 0. Otherwise, if (0.0000213444a - 1.48) = 0, solving for a gives:

0.0000213444a = 1.48

a ≈ 69.307 m/s²

Therefore, the magnitude of the average acceleration a of the ball while it is in contact with the floor is approximately 69.307 m/s².

To find the magnitude of the average acceleration of the golf ball while it is in contact with the floor, you can use the kinematic equation:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

In this case, we are given the initial height (h1) as 1.80 m, and the final height (h2) as 1.06 m. We can use the equation for the height of an object in free fall:

h = ut + (1/2)at^2

Since the ball is released from rest, the initial velocity (u) is 0. Let's consider the time it takes for the golf ball to fall from the initial height to the floor as t1, and the time it takes for the ball to rise from the floor to the final height as t2.

Using the equation for the height h1:

h1 = (1/2)at1^2

Solving for t1:

2h1 / a = t1^2
t1 = sqrt(2h1 / a)

Using the equation for the height h2:

h2 = (1/2)at2^2

Solving for t2:

2h2 / a = t2^2
t2 = sqrt(2h2 / a)

Since the total time the ball is in contact with the floor is given as 4.62 ms, we have:

t1 + t2 = 4.62 ms

Substituting the values for t1 and t2:

sqrt(2h1 / a) + sqrt(2h2 / a) = 4.62 ms

Now, you can use algebraic techniques to solve this equation for the value of a.

the velocity when the ball first contacts the floor is:

V1 = sqrt(2*g*s)
V1= sqrt(2 * 9.8(m/s^s) * 1.8m)

The velocity when the ball just leaves (rebounds) works the same way except for the distance sign
V2 = sqrt(2 * 9.8(m/s^2) * (-1.06m))

V1 - V2= delta V (you will be subtracting a negative number remember for V2.

It is given that delta t is 4.62ms

acceleration = (delta v) / (delta t)

Don't forget to change the ms time to seconds.