Question

Make a prediction of vertical velocity versus time (vy vs. t) for a ball in free fall from bouncing vertically on a table. You want to predict the entire motion from right after the first contact with the table (on its way up), to the top of the motion, and back down to the table at the instant before the ball hits the table for the second time (on its way down, but hasn’t hit yet). Graph only predictions for that one bounce, not multiple bounces.

(a) Sketch a graph of vertical velocity versus time (vy vs. t) for a ball in free-fall.

(b) Explain your prediction – how is the physical motion of the ball reflected in the graph you drew? Is the equation for this motion linear or quadratic? Also, discuss how the graph shows the ball changing direction.

Answer 1

Answer 2

(a) Graph Sketch:

Below is a description of the vₓ vs. t graph for one complete bounce (from first contact upward to just before second impact):

 

vₓ (m/s) ▲
          │
   +v₀    │      /\
          │     /  \
          │    /    \
  0 ------┼-----------► t (s)
          │  \      /
          │   \    /
   -v₀    │    \/ 
          │

• Key Features:

• At t=0: Ball leaves the table with initial upward velocity (+v₀).

• Rising Phase (0 → t₁): Velocity decreases linearly due to gravity (slope = −g = −9.8 m/s²).

• At Peak (t₁): Velocity momentarily = 0 (ball changes direction).

• Falling Phase (t₁ → t₂): Velocity increases negatively (slope = −g) until just before impact (−v₀).

• 
  

(b) Explanation:

1. Physical Motion Reflected in Graph:

• Linear Slope: The constant slope (−g) reflects uniform acceleration due to gravity.

• Direction Change: The vₓ=0 point marks the peak height where the ball stops rising and begins falling.

• Symmetry: If no energy is lost, |+v₀| = |−v₀| (elastic collision).

• 
  

2. Equation of Motion:

• Velocity is Linear with Time:

  $$v_y(t)=v_0-gt$$

• Position is Quadratic:

  $$y(t)=v_{0}t-\frac{1}{2}g{t}^2$$

3. How the Graph Shows Direction Change:

• Positive vₓ (↑): Ball moves upward.

• Negative vₓ (↓): Ball moves downward.

• Zero Crossing: Instantaneous stop at the peak.

• 
  

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Key Notes:

• Assumptions: Negligible air resistance, perfectly elastic collision (no energy loss).

• 
  

• Real-World Adjustments: With energy loss, |−v₀| < |+v₀| (damped bounces).

• 
  

 Answer:
The vₓ vs. t graph is piecewise linear with a constant slope of −g, capturing the ball’s deceleration, direction change, and re-acceleration downward.