A 62.7-kg basketball player jumps vertically and leaves the floor with a velocity of 1.77 m/s upward. (c)
What is the total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump?

To solve for the total average force exerted by the floor on the player during the jump, we can use the concept of impulse and Newton's Second Law. The impulse applied to the player is equal to the change in momentum, and the force can be calculated using the following steps:

### 1. **Find the player's change in momentum:**
   The change in momentum is the difference between the final momentum (when the player leaves the ground) and the initial momentum (when the player is stationary).

   \[
   \Delta p = m \cdot \Delta v = m \cdot v_f - m \cdot v_i
   \]
   Here:
   - \( m = 62.7 \, \text{kg} \) (mass of the player),
   - \( v_f = 1.77 \, \text{m/s} \) (final velocity),
   - \( v_i = 0 \, \text{m/s} \) (initial velocity, since the player starts from rest).

   So, the change in momentum is:

   \[
   \Delta p = 62.7 \times (1.77 - 0) = 111.979 \, \text{kg} \cdot \text{m/s}
   \]

### 2. **Apply the impulse-momentum theorem:**
   Impulse is also defined as the force exerted multiplied by the time of contact. Using the formula for impulse:

   \[
   F \cdot \Delta t = \Delta p
   \]
   Rearranging to solve for \( F \), we get:

   \[
   F = \frac{\Delta p}{\Delta t}
   \]

   Where:
   - \( \Delta p = 111.979 \, \text{kg} \cdot \text{m/s} \),
   - \( \Delta t = 0.450 \, \text{s} \) (time of contact with the floor).

   Now, calculating the average force:

   \[
   F = \frac{111.979}{0.450} = 248.84 \, \text{N}
   \]

### 3. **Include gravitational force:**
   The total force exerted by the floor must also counteract the gravitational force acting on the player. The gravitational force is given by:

   \[
   F_{\text{gravity}} = m \cdot g
   \]
   Where:
   - \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity),
   - \( m = 62.7 \, \text{kg} \).

   So, the gravitational force is:

   \[
   F_{\text{gravity}} = 62.7 \times 9.81 = 614.19 \, \text{N}
   \]

### 4. **Total average force exerted by the floor:**
   The total force exerted by the floor is the sum of the force required to produce the jump and the force required to support the player's weight:

   \[
   F_{\text{total}} = F + F_{\text{gravity}} = 248.84 + 614.19 = 863.03 \, \text{N}
   \]

### Final Answer:
The total average force exerted by the floor on the player during the jump is **863.03 N**.

Would you like further details or clarifications?

### Related Questions:
1. How would the force change if the player had a mass of 70 kg instead of 62.7 kg?
2. What would be the impact on the force if the contact time with the floor was shorter, say 0.30 s?
3. How does the final velocity affect the force exerted by the floor on the player?
4. How would the calculated force change if the player were jumping on a planet with lower gravity?
5. What would happen to the force if the player increased their jump velocity to 2.5 m/s?

### Tip:
The longer the time of contact with the ground, the smaller the force required to achieve the same change in momentum.

A 62.7-kg basketball player jumps vertically and leaves the floor with a velocity of 1.77 m/s upward. What is the total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump?

Learn how to calculate the total average force exerted by the floor on a 62.7-kg basketball player who jumps vertically with a velocity of 1.77 m/s and a contact time of 0.450 s. This solution applies the impulse-momentum theorem, Newton's second law, and includes the effect of gravitational force. Suitable for students in Grades 10-12.

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