Let's go through each part of the problem step-by-step and verify the calculations.

Problem 20: Chris wants to dunk a basketball.

Part (a) - Calculating Jump Height

• Given values:

  • $v_0 = 3.724 \, \text{m/s}$ (initial velocity)
  • $v_f = 0 \, \text{m/s}$ (velocity at the highest point)
  • $a = -9.8 \, \text{m/s}^2$ (acceleration due to gravity)
  • $t = 0.38 \, \text{s}$ (time to reach the highest point)

• Formula: We use the kinematic equation: $y = \frac{v_f^2 - v_0^2}{2a}$

• Calculation: $y = \frac{0^2 - (3.724)^2}{2 \times (-9.8)} = \frac{-13.864}{-19.6} = 0.708 \, \text{m}$

Chris jumps 0.708 meters high.

Part (b) - Calculating Total Hang Time

• Explanation: The time to reach the highest point is $0.38 \, \text{s}$. Since the motion is symmetric, the total time (up and down) is: $\text{Total time} = 0.38 \times 2 = 0.76 \, \text{s}$

Chris's total hang time is 0.76 seconds.

Part (c) - Can Chris Dunk?

• Given:

  • The basketball hoop height is $3.05 \, \text{m}$.
  • Chris's reach with his arms extended is $2.29 \, \text{m}$.

• Calculation: $\text{Max height Chris can reach} = 2.29 + 0.708 = 2.998 \, \text{m}$

Since 2.998 m < 3.05 m, Chris cannot dunk.

Part (d) - Calculating Gravity on Planet Doucettene

• Given values:

  • $y = 3.19 \, \text{m}$ (height Chris reached on Doucettene)
  • $v_0 = 3.724 \, \text{m/s}$
  • $v_f = 0 \, \text{m/s}$

• Formula: We rearrange the same kinematic equation used in part (a): $a = \frac{v_f^2 - v_0^2}{2y}$

• Calculation: $a = \frac{0^2 - (3.724)^2}{2 \times 3.19} = \frac{-13.864}{6.38} = -2.17 \, \text{m/s}^2$

The acceleration due to gravity on Doucettene is -2.17 m/s².

Would you like a breakdown of any part of these solutions?

Here are some follow-up questions you might find helpful:

1. Why is gravity's value negative in the equations?
2. What would Chris's jump height be if he jumped with a different initial velocity?
3. How would Doucettene’s lower gravity affect Chris’s hang time?
4. Could Chris achieve the necessary height with a slight increase in jump speed on Earth?
5. Why do we assume that air resistance is negligible in these calculations?

Tip: Always double-check unit consistency across equations, as it helps prevent errors.