The elevator starts from rest at the level of the earth's surface and accelerates with a constant acceleration a
. When time has passed t1=9.0 s
, an object is dropped from the opening at the bottom of the elevator. The object hits the ground t2=4.2 s
 after the fall. Air resistance is not taken into account.
a) What is the magnitude of the elevator's acceleration a?
b) What was the lift's height h when the object was dropped?

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.

The elevator starts from rest at the level of the earth's surface and accelerates with a constant acceleration a. When time has passed t1 = 9.0 s, an object is dropped from the opening at the bottom of the elevator. The object hits the ground t2 = 4.2 s after the fall. Air resistance is not taken into account. a) What is the magnitude of the elevator's acceleration a? b) What was the lift's height h when the object was dropped?

Learn how to calculate the acceleration and height of an elevator using kinematic equations in this physics problem. We cover key concepts such as constant acceleration, free fall, and motion equations. Suitable for Grades 11-12 or introductory college physics students.

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