To solve this problem, we'll use the equations of motion under constant acceleration. Let's break it down into two parts:

1. Determine the acceleration $a$ of the elevator.
2. Find the height $h$ of the elevator when the object was dropped.

Given Information:

• Time during which the elevator accelerates: $t_1 = 9.0$ seconds.
• Time taken by the object to hit the ground after being dropped: $t_2 = 4.2$ seconds.
• The object is dropped at time $t_1$.
• Acceleration due to gravity: $g = 9.8 \, \text{m/s}^2$.

Step-by-Step Solution:

Part (a): Determine the acceleration $a$ of the elevator

1. Find the velocity of the elevator when the object is dropped.

The elevator starts from rest and accelerates with a constant acceleration $a$. The velocity $v$ of the elevator at time $t_1$ is given by:

$v = a \cdot t_1$

2. Find the distance traveled by the elevator when the object is dropped.

The distance $h$ traveled by the elevator when the object is dropped at time $t_1$ can be calculated using the formula:

$h = \frac{1}{2} a t_1^2$

3. Consider the motion of the dropped object.

When the object is dropped, it has the initial velocity $v = a \cdot t_1$ (the same as the velocity of the elevator at time $t_1$). It will then move under the influence of gravity alone.

The equation for the height $h$ of the object after time $t_2$ is:

$h = v \cdot t_2 - \frac{1}{2} g t_2^2$

Since we know that the object reaches the ground (i.e., $h = 0$) at time $t_2$, we have:

$0 = (a \cdot t_1) \cdot t_2 - \frac{1}{2} g t_2^2$

Now, solve this equation for the acceleration $a$.

$a \cdot t_1 \cdot t_2 = \frac{1}{2} g t_2^2$

$a = \frac{\frac{1}{2} g t_2^2}{t_1 \cdot t_2}$

$a = \frac{g t_2}{2 t_1}$

Substitute the given values $g = 9.8 \, \text{m/s}^2$, $t_2 = 4.2 \, \text{s}$, and $t_1 = 9.0 \, \text{s}$:

$a = \frac{9.8 \times 4.2}{2 \times 9.0}$

Now let's calculate this value:

Part (b): Find the height $h$ of the elevator when the object was dropped

1. Use the height formula:

$h = \frac{1}{2} a t_1^2$

We will substitute the value of $a$ obtained from part (a) and $t_1 = 9.0 \, \text{s}$ to find $h$.

Let's calculate these values.The magnitude of the elevator's acceleration is approximately $a = 2.29 \, \text{m/s}^2$.

The height of the elevator when the object was dropped is approximately $h = 92.61 \, \text{m}$.

Would you like further details on any part of this solution? Here are some related questions you might consider:

1. What if the elevator was decelerating instead? How would that affect the results?
2. How would the result change if air resistance were considered?
3. Can we derive a general formula for an object dropped from a moving elevator?
4. What if the elevator was moving upwards with a constant speed when the object was dropped?
5. How would the acceleration of gravity change on a different planet, and what impact would that have on this problem?

Tip: When solving motion problems, always check if the direction of acceleration and initial velocity align to apply the correct signs in equations.