Vertical Motion

At a construction site, a person is standing on an elevator that is initially at rest.  The elevator does not have a ceiling. The person throws a ball straight up at a speed of 13.7 m/s.  The ball leaves the person’s hand at a height of 1.85 m from the elevator floor.  At the instant the ball leaves the person's hand, the elevator starts moving downwards at a constant speed of 0.815 m/s.

a) Calculate the height of the ball above the elevator floor when the ball reaches its maximum height.

Answer:  12.56 m

After reaching its maximum height, the ball falls down to the elevator floor.

b) Calculate the time when the ball hits the elevator floor.

Answer: s

c) Calculate the speed with which the ball hits the elevator floor.

Answer: m/s

d) Calculate the distance travelled by the elevator floor when the ball hits it.

Answer: m

### Part (a): Maximum Height of the Ball Above the Elevator Floor

To find the height of the ball above the elevator floor when it reaches its maximum height, we first analyze the motion of the ball in two parts: the vertical motion of the ball and the downward movement of the elevator.

1. **Vertical Motion of the Ball:**
   The ball is thrown vertically upwards with an initial velocity \( v_0 = 13.7 \, \text{m/s} \). The acceleration due to gravity acts downward with \( g = 9.8 \, \text{m/s}^2 \). The maximum height occurs when the ball’s velocity becomes zero.

   Using the kinematic equation for vertical motion:
   \[
   v = v_0 - g t
   \]
   At the maximum height, \( v = 0 \), so:
   \[
   0 = 13.7 - 9.8 t
   \]
   Solving for \( t \), we get the time to reach maximum height:
   \[
   t = \frac{13.7}{9.8} \approx 1.4 \, \text{seconds}
   \]

2. **Maximum Height of the Ball Above Initial Position:**
   Using the kinematic equation for position:
   \[
   h = v_0 t - \frac{1}{2} g t^2
   \]
   Substituting values:
   \[
   h = 13.7 \times 1.4 - \frac{1}{2} \times 9.8 \times (1.4)^2
   \]
   \[
   h \approx 19.18 - 9.6 \approx 9.58 \, \text{m}
   \]
   So, the ball rises \( 9.58 \, \text{m} \) above the initial position of 1.85 m. Therefore, the total height of the ball above the elevator floor at its maximum height is:
   \[
   \text{Total height} = 9.58 + 1.85 = 11.43 \, \text{m}
   \]

3. **Height of the Elevator Floor:**
   In the same 1.4 seconds, the elevator moves down at a constant speed of \( 0.815 \, \text{m/s} \). The distance traveled by the elevator is:
   \[
   d_{\text{elevator}} = v_{\text{elevator}} \times t = 0.815 \times 1.4 \approx 1.14 \, \text{m}
   \]

4. **Height of the Ball Above the Elevator Floor:**
   Therefore, the height of the ball above the elevator floor at maximum height is:
   \[
   11.43 + 1.14 = 12.56 \, \text{m}
   \]

This matches the provided answer, \( 12.56 \, \text{m} \).

---

### Part (b): Time When the Ball Hits the Elevator Floor

To calculate the time when the ball hits the elevator floor, we need to track the motion of both the ball and the elevator.

1. **Total Displacement of the Ball:**
   The ball initially rises to a maximum height and then falls. The total distance the ball travels down to the elevator floor is the height at its peak (11.43 m) plus the 1.85 m initial height:
   \[
   \text{Total displacement of the ball} = 11.43 + 1.85 = 13.28 \, \text{m}
   \]

2. **Relative Motion of the Ball to the Elevator:**
   The ball falls while the elevator continues to move downward. The relative distance between them must be calculated as a function of time. We will solve this system of equations for the time when they meet.

We'll break this into two steps for clarity:

- **Step 1:** Calculate the total time it takes for the ball to fall from its maximum height.
- **Step 2:** Include the downward movement of the elevator and solve

At a construction site, a person is standing on an elevator that is initially at rest. The elevator does not have a ceiling. The person throws a ball straight up at a speed of 13.7 m/s. The ball leaves the person’s hand at a height of 1.85 m from the elevator floor. At the instant the ball leaves the person's hand, the elevator starts moving downwards at a constant speed of 0.815 m/s.

a) Calculate the height of the ball above the elevator floor when the ball reaches its maximum height.

b) Calculate the time when the ball hits the elevator floor.

c) Calculate the speed with which the ball hits the elevator floor.

d) Calculate the distance travelled by the elevator floor when the ball hits it.

Solve a vertical motion problem where a ball is thrown upwards at 13.7 m/s from a height of 1.85 m, while the elevator moves downwards at a constant speed of 0.815 m/s. Learn to calculate the maximum height, time of impact, and speed of the ball when it hits the elevator floor. Suitable for Grades 9-12.

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