A ball is thrown vertically upward from the 12 m level with an initial velocity of `18 m//s.` At the same instant an open platform elevator passes the 5 m level, moving upward with a constant velocity of `2 m//s.` Determine (`g = 9.8 m//s^2` )
(a) when and where the ball will meet the elevator,
(b) the relative velocity of the ball with respect to the elevator when the ball hits the elevator.

Explanation

To solve the problem step by step, we will analyze the motion of both the ball and the elevator separately and then find the time and position where they meet.

### Given Data:
- Initial height of the ball, \( h_b = 12 \, \text{m} \)
- Initial velocity of the ball, \( u_b = 18 \, \text{m/s} \) (upward)
- Initial height of the elevator, \( h_e = 5 \, \text{m} \)
- Velocity of the elevator, \( v_e = 2 \, \text{m/s} \) (upward)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 1: Write the equations of motion for both the ball and the elevator.

**For the ball:**
Using the equation of motion:
\[ s_b = h_b + u_b t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ s_b = 12 + 18t - 4.9t^2 \]

**For the elevator:**
Since the elevator moves with constant velocity:
\[ s_e = h_e + v_e t \]
Substituting the values:
\[ s_e = 5 + 2t \]

### Step 2: Set the equations equal to find when they meet.

At the point of meeting:
\[ s_b = s_e \]
Thus,
\[ 12 + 18t - 4.9t^2 = 5 + 2t \]

### Step 3: Rearrange the equation.

Rearranging gives:
\[ -4.9t^2 + 18t - 2t + 12 - 5 = 0 \]
\[ -4.9t^2 + 16t + 7 = 0 \]

### Step 4: Solve the quadratic equation.

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = -4.9, b = 16, c = 7 \).

Calculating the discriminant:
\[ D = b^2 - 4ac = 16^2 - 4 \times (-4.9) \times 7 \]
\[ D = 256 + 137.2 = 393.2 \]

Now substituting into the quadratic formula:
\[ t = \frac{-16 \pm \sqrt{393.2}}{2 \times -4.9} \]
Calculating \( \sqrt{393.2} \approx 19.83 \):
\[ t = \frac{-16 \pm 19.83}{-9.8} \]

Calculating the two possible values for \( t \):
1. \( t = \frac{3.83}{-9.8} \) (not valid as time cannot be negative)
2. \( t = \frac{-35.83}{-9.8} \approx 3.65 \, \text{s} \)

### Step 5: Find the height at which they meet.

Substituting \( t = 3.65 \) into the elevator's equation:
\[ s_e = 5 + 2(3.65) = 5 + 7.3 = 12.3 \, \text{m} \]

### Step 6: Calculate the relative velocity of the ball with respect to the elevator.

The velocity of the ball at \( t = 3.65 \) is given by:
\[ v_b = u_b - gt = 18 - 9.8 \times 3.65 \]
Calculating:
\[ v_b = 18 - 35.77 \approx -17.77 \, \text{m/s} \]

The velocity of the elevator is constant:
\[ v_e = 2 \, \text{m/s} \]

Thus, the relative velocity of the ball with respect to the elevator:
\[ v_{rel} = v_b - v_e = -17.77 - 2 = -19.77 \, \text{m/s} \]

### Final Answers:
(a) The ball meets the elevator at \( t = 3.65 \, \text{s} \) at a height of \( 12.3 \, \text{m} \).

(b) The relative velocity of the ball with respect to the elevator when the ball hits the elevator is \( 19.77 \, \text{m/s} \) downward.