A tennis ball of mass 57.0 g is held just above a basketball of mass 608 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.36 m, as shown in the figure below. (a) Find the magnitude of the downward velocity with which the basketball reaches the ground. (b) Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound? Step 1 (a) When released, both balls start from rest and fall freely under the influence of gravity for 1.36 m before the basketball touches the ground. Their downward speed just before the basketball hits the ground is found from Voy - 2a Ay. In this problem we have that vy is the speed at impact with the ground of the basketball v. Solving for v; from the equation above, we have Vy m/s. V₁ = V 0 + 2(-9.80 m/s²) + 2a Ay = Voy n/s²)(- Note that both the acceleration due to gravity and the vertical displacement through which the balls fall are negative.

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**Physics Simulation: Double Ball Drop**

**Introduction:**
A tennis ball of mass 57.0 g is held just above a basketball of mass 608 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.36 m.

**Goal:**
(a) Calculate the magnitude of the downward velocity with which the basketball reaches the ground.

(b) Assume an elastic collision with the ground that instantly reverses the velocity of the basketball, while the tennis ball is still falling. Then, the two balls collide elastically in the air. Determine to what height the tennis ball rebounds.

---

**Step 1: Calculate Velocity Upon Impact**

**Procedure:**
(a) Upon release, both balls fall freely due to gravity for 1.36 m before the basketball touches the ground. The equation for downward speed just before the basketball hits the ground is:

\[ v_y^2 = v_{0y}^2 + 2a_y \Delta y. \]

- In this case, \( v_y \) is the speed at impact for the basketball.
- Solve for \( v_y \):

\[ v_i = \sqrt{v_{0y}^2 + 2a_y \Delta y} = \sqrt{0 + 2(-9.80 \, \text{m/s}^2)(-1.36 \, \text{m})} = \text{______} \, \text{m/s}. \]

**Note:**
Both the acceleration due to gravity and the vertical displacement are taken as negative.

**Diagram:**
The diagram shows a tennis ball positioned directly above a basketball, with arrows indicating the direction of the force of gravity acting on both balls as they fall towards the ground.
Transcribed Image Text:**Physics Simulation: Double Ball Drop** **Introduction:** A tennis ball of mass 57.0 g is held just above a basketball of mass 608 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.36 m. **Goal:** (a) Calculate the magnitude of the downward velocity with which the basketball reaches the ground. (b) Assume an elastic collision with the ground that instantly reverses the velocity of the basketball, while the tennis ball is still falling. Then, the two balls collide elastically in the air. Determine to what height the tennis ball rebounds. --- **Step 1: Calculate Velocity Upon Impact** **Procedure:** (a) Upon release, both balls fall freely due to gravity for 1.36 m before the basketball touches the ground. The equation for downward speed just before the basketball hits the ground is: \[ v_y^2 = v_{0y}^2 + 2a_y \Delta y. \] - In this case, \( v_y \) is the speed at impact for the basketball. - Solve for \( v_y \): \[ v_i = \sqrt{v_{0y}^2 + 2a_y \Delta y} = \sqrt{0 + 2(-9.80 \, \text{m/s}^2)(-1.36 \, \text{m})} = \text{______} \, \text{m/s}. \] **Note:** Both the acceleration due to gravity and the vertical displacement are taken as negative. **Diagram:** The diagram shows a tennis ball positioned directly above a basketball, with arrows indicating the direction of the force of gravity acting on both balls as they fall towards the ground.
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