An object shot straight up rises for 7.5 s before it reaches its maximum height. A second object falling from rest takes 7.5 s to reach the ground. Compare the displacements of the two objects during this time interval.

(1) Both objects traveled the same distance. The object that is shot straight upward rises to the same height from which the other object fell.
(2) The objects traveled different distances. The object that is shot straight upward rises to a height lower than the height from which the other object fell.
(3) The objects traveled different distances. The object that is shot straight upward rises to a height higher than the height from which the other object fell.

To compare the displacements of the two objects, we can analyze their motions using the principles of kinematics and symmetry of free-fall motion under uniform gravity.

### Object 1 (shot straight upward):
- The object rises for 7.5 seconds before reaching its maximum height.
- The initial velocity is positive (upward), and the acceleration is \( g = 9.8 \, \text{m/s}^2 \) downward.
- At maximum height, its velocity becomes zero after 7.5 seconds.

Using the kinematic equation for displacement (\( d \)):

\[
d_1 = v_0 t + \frac{1}{2} a t^2
\]

Where:
- \( v_0 \) is the initial velocity,
- \( a = -9.8 \, \text{m/s}^2 \) (negative because gravity acts downward),
- \( t = 7.5 \, \text{seconds} \).

Since we are considering the object going up and stopping at its maximum height, we can use another useful equation to determine the maximum height:

\[
v = v_0 - g t
\]

At maximum height, \( v = 0 \), so:

\[
0 = v_0 - g t
\]

Solving for \( v_0 \):

\[
v_0 = g t = 9.8 \times 7.5 = 73.5 \, \text{m/s}
\]

Now, use this to find the displacement:

\[
d_1 = v_0 t - \frac{1}{2} g t^2 = 73.5 \times 7.5 - \frac{1}{2} \times 9.8 \times (7.5)^2
\]

\[
d_1 = 551.25 - 275.63 = 275.62 \, \text{m}
\]

So the object rises to a maximum height of \( 275.62 \, \text{m} \).

### Object 2 (falling from rest):
- The object is dropped from rest and falls for 7.5 seconds.
- Initial velocity \( v_0 = 0 \),
- The acceleration is \( g = 9.8 \, \text{m/s}^2 \) downward.

Using the same kinematic equation:

\[
d_2 = v_0 t + \frac{1}{2} g t^2
\]

Since \( v_0 = 0 \):

\[
d_2 = \frac{1}{2} g t^2 = \frac{1}{2} \times 9.8 \times (7.5)^2
\]

\[
d_2 = \frac{1}{2} \times 9.8 \times 56.25 = 275.63 \, \text{m}
\]

### Comparison:
Both objects traveled the same distance of \( 275.63 \, \text{m} \). The object that is shot upward rises to the same height from which the second object falls.

Thus, the correct answer is:

**(1) Both objects traveled the same distance. The object that is shot straight upward rises to the same height from which the other object fell.**

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Would you like more details on any step? Here are 5 questions to expand the topic:

1. How would air resistance affect the motion of both objects?
2. How can we calculate the total time of flight for the upward-shooting object?
3. What are the energy transformations involved in the upward motion and free fall?
4. How would the scenario change if the acceleration due to gravity were different?
5. What would happen if the object were thrown at an angle instead of straight upward?

**Tip**: In free-fall problems, symmetry is key. The time to rise to the maximum height equals the time to fall back down from that height.

An object shot straight up rises for 7.5 s before it reaches its maximum height. A second object falling from rest takes 7.5 s to reach the ground. Compare the displacements of the two objects during this time interval.

Explore how to compare the displacements of two objects: one shot straight up and the other falling from rest, both in free fall for 7.5 seconds. This detailed solution includes kinematic equations, symmetry principles, and motion under gravity. Suitable for Grades 9-12.

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