A watermelon is dropped from the edge of the roof of a building and falls to the ground. You are standing on the sidewalk and see the watermelon falling when it is 30.0 m
 above the ground. Then 1.50 s
 after you first spot it, the watermelon lands at your feet. Neglect air resistance.
Part A
What is the height of the building?
Express your answer with the appropriate units.
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H
  =


We can determine the height of the building using the kinematic equations of motion. Here's how we approach the problem step by step:

### Given:
- You observe the watermelon at a height of 30.0 m above the ground.
- The time between seeing the watermelon and it hitting the ground is \( t = 1.50 \, \text{s} \).
- The acceleration due to gravity \( g = 9.80 \, \text{m/s}^2 \) (since air resistance is neglected).
- The initial velocity \( v_0 = 0 \, \text{m/s} \) (since the watermelon was dropped).

We can first find the velocity of the watermelon after 1.50 s when it hits the ground, then calculate the height from which it was dropped.

### Step 1: Using the kinematic equation to find the total height the watermelon fell from the point you spotted it (30.0 m above the ground):
\[
h = v_0 t + \frac{1}{2} g t^2
\]
Since the watermelon is falling from rest, \( v_0 = 0 \), so the equation simplifies to:
\[
h = \frac{1}{2} g t^2
\]

Substituting the values:
\[
h = \frac{1}{2} (9.80 \, \text{m/s}^2)(1.50 \, \text{s})^2
\]

Let’s calculate the distance \( h \).The watermelon falls a distance of \( 11.025 \, \text{m} \) during the 1.50 seconds.

### Step 2: Total height of the building
Since the watermelon was 30.0 m above the ground when you first saw it and then fell an additional 11.025 m, the total height of the building is:

\[
H = 30.0 \, \text{m} + 11.025 \, \text{m} = 41.025 \, \text{m}
\]

Thus, the height of the building is approximately \( 41.0 \, \text{m} \).

---

Would you like further details or explanations on any part of this? Here are some related questions to consider:

1. What is the velocity of the watermelon when it hits the ground?
2. How much time did the watermelon take to fall the entire distance from the building?
3. What if air resistance was not neglected? How would the motion change?
4. How would the result change if the building was on a hill with a sloped ground?
5. Can we use energy conservation principles to solve this problem?

**Tip:** When solving free-fall problems, always start by identifying whether the object was dropped or thrown, as this affects the initial velocity.

A watermelon is dropped from the edge of the roof of a building and falls to the ground. You are standing on the sidewalk and see the watermelon falling when it is 30.0 m above the ground. Then 1.50 s after you first spot it, the watermelon lands at your feet. Neglect air resistance. Part A: What is the height of the building?

Learn how to calculate the height of a building using kinematic equations of motion. This problem involves free-fall motion, gravity, and the observation of a watermelon falling from a height of 30.0 m to the ground in 1.50 seconds. Suitable for students in Grades 10-12.

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