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.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:

To calculate when the ball hits the elevator floor, we need to take into account the motion of both the ball and the elevator. Here's the step-by-step process:

### Step 1: Establish the motion of the ball
We are given that the ball is thrown upwards with an initial speed \( v_0 = 13.7 \, \text{m/s} \) from a height of \( h_0 = 1.85 \, \text{m} \) above the elevator floor.

The motion of the ball follows the equations of uniformly accelerated motion, where the acceleration is due to gravity:
\[
y_b(t) = h_0 + v_0 t - \frac{1}{2} g t^2
\]
where:
- \( y_b(t) \) is the height of the ball relative to the elevator floor after time \( t \),
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( h_0 = 1.85 \, \text{m} \),
- \( v_0 = 13.7 \, \text{m/s} \).

Thus, the equation for the ball's height above the elevator floor becomes:
\[
y_b(t) = 1.85 + 13.7t - 4.9t^2
\]

### Step 2: Establish the motion of the elevator
The elevator moves downwards with a constant speed of \( v_e = 0.815 \, \text{m/s} \). So, the position of the elevator floor after time \( t \) is:
\[
y_e(t) = -0.815 t
\]

### Step 3: Calculate when the ball hits the elevator floor
The ball hits the elevator floor when their positions are equal, i.e., \( y_b(t) = y_e(t) \). Thus, we set:
\[
1.85 + 13.7t - 4.9t^2 = -0.815t
\]
Simplify this equation:
\[
1.85 + 13.7t - 4.9t^2 + 0.815t = 0
\]
\[
1.85 + 14.515t - 4.9t^2 = 0
\]
Now, rearrange it into a standard quadratic form:
\[
-4.9t^2 + 14.515t + 1.85 = 0
\]

### Step 4: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = -4.9 \), \( b = 14.515 \), and \( c = 1.85 \).

Now let's compute the solution.The solutions for the time are:

\[
t_1 \approx 3.13 \, \text{seconds}
\]
\[
t_2 \approx -0.12 \, \text{seconds}
\]

Since time cannot be negative, the valid solution is:

\[
t = 3.13 \, \text{seconds}
\]

Thus, the ball hits the elevator floor approximately **3.13 seconds** after it is thrown.

Would you like further details on any step? Here are five questions that could expand on this topic:

1. How is the motion of the ball described using kinematic equations?
2. What is the effect of gravity on the ball's motion after it's thrown upwards?
3. How does the elevator's constant downward speed influence the time calculation?
4. What are the steps involved in solving quadratic equations in physics problems?
5. How would the time change if the elevator was moving upwards instead of downwards?

**Tip:** When solving physics problems with multiple objects in motion, always set up their individual equations first and then solve for the point where they meet or interact.

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. After reaching its maximum height, the ball falls down to the elevator floor. Calculate the time when the ball hits the elevator floor.

Solve the kinematic problem of a ball thrown upwards from a moving elevator at 13.7 m/s. Learn how to use quadratic equations and motion formulas to calculate the time when the ball hits the elevator floor. Key concepts include kinematics and the quadratic formula. Suitable for Grades 10-12.

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