A 116.0 g ball is dropped from a height of 2.75 m, bounces on a hard floor, and rebounds to a height of 1.82 m. The impulse received from the floor is shown below. What maximum force does the floor exert on the ball if it is exerted for 7.00 ms?

force*timeimpact=changemomentum=massball(vf-vi)

calculate vi, vf from the heights...remember that they are OPPOSITE signs.
calculate force.

when calculating the velocities from the heights, what time interval would you use

and would the force be positive or negative

To find the maximum force exerted by the floor on the ball, we can use the concept of impulse. Impulse is given by the product of force and time, and it is equal to the change in momentum of an object.

The change in momentum is the difference between the initial momentum and the final momentum. In this case, since the ball is dropped from rest, the initial momentum is zero. The final momentum can be calculated using the mass of the ball and its velocity just after bouncing.

The impulse received from the floor is given in the question as 0.378 Ns. We can equate this impulse to the change in momentum using the equation:

Impulse = Change in momentum

0.378 Ns = Final momentum - Initial momentum

Since the initial momentum is zero, we can simplify the equation to:

0.378 Ns = Final momentum

To find the final momentum of the ball, we need to calculate its velocity just after bouncing. We can use the concept of conservation of mechanical energy. The potential energy of the ball at its initial height is equal to the kinetic energy of the ball just after bouncing.

The potential energy is given by:
P.E = mgh

Where:
m = mass of the ball = 116.0 g = 0.116 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the ball = 2.75 m

P.E = 0.116 kg * 9.8 m/s^2 * 2.75 m = 3.0612 J

The kinetic energy is given by:
K.E = (1/2)mv^2

Where:
m = mass of the ball = 116.0 g = 0.116 kg
v = velocity of the ball just after bouncing

Since the ball starts from rest, the initial velocity is zero, so we can simplify the equation to:
K.E = (1/2)mv^2 = (1/2) * 0.116 kg * v^2

From the conservation of mechanical energy:
P.E = K.E
3.0612 J = (1/2) * 0.116 kg * v^2
v^2 = (3.0612 J * 2) / 0.116 kg
v^2 = 52.671 J/kg
v = √(52.671 J/kg) = 7.255 m/s (approx.)

Now that we know the velocity just after bouncing, we can calculate the final momentum using the equation:

Final momentum = mass * velocity
Final momentum = 0.116 kg * 7.255 m/s = 0.8423 kg·m/s (approx.)

Now we can substitute the final momentum into the equation for impulse to find the maximum force exerted by the floor:

0.378 Ns = 0.8423 kg·m/s - 0

Simplifying the equation, we find:
0.378 Ns = 0.8423 kg·m/s
Dividing both sides of the equation by the time in seconds (7.00 ms = 0.007 s) gives us:
Maximum force = (0.8423 kg·m/s) / (0.007 s) = 120.33 N (to three significant figures)

Therefore, the maximum force exerted by the floor on the ball is approximately 120.33 N.