https://file.aiquickdraw.com/mathbot/file/1727311681450zy2rvlog.jpg

Let's break down the problem presented in the image:

- **Given Information:**
  - A ball is launched at an angle of **45°** above the horizontal.
  - It reaches a **maximum height** of **40.0 m**.

You are asked to find the **range** of the projectile, which is the horizontal distance the ball travels before hitting the ground.

### Step-by-Step Solution:

#### 1. **Find the initial velocity using maximum height:**

The formula for maximum height \( H \) in projectile motion is:

\[
H = \frac{v_0^2 \sin^2 \theta}{2g}
\]

Where:
- \( H \) = maximum height (40.0 m)
- \( v_0 \) = initial velocity (unknown)
- \( \theta \) = angle of launch (45°)
- \( g \) = acceleration due to gravity (9.8 m/s²)

Rearranging to solve for \( v_0 \):

\[
v_0^2 = \frac{2gH}{\sin^2 \theta}
\]

Substitute the known values:

\[
v_0^2 = \frac{2 \times 9.8 \times 40.0}{\sin^2 45^\circ}
\]

Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), we can now calculate the value of \( v_0 \).

#### 2. **Calculate the Range:**

The range \( R \) of a projectile is given by the formula:

\[
R = \frac{v_0^2 \sin 2\theta}{g}
\]

Since \( \sin 90^\circ = 1 \) (because \( 2 \times 45^\circ = 90^\circ \)), this simplifies to:

\[
R = \frac{v_0^2}{g}
\]

Now, I'll calculate the exact values for both \( v_0 \) and \( R \).The initial velocity \( v_0 \) is approximately **39.60 m/s**, and the range \( R \) of the projectile is approximately **160.0 meters**.

Thus, the correct answer for the range is **160.0 m**, not **40.0 m** as selected.

Do you need any further clarifications or details? Here are some related questions to consider:

1. How would the range change if the angle of projection were increased or decreased?
2. What is the total time of flight for this projectile?
3. How would air resistance affect the range of the projectile?
4. Can you derive the range formula from the basic kinematic equations?
5. How does the initial velocity influence both the maximum height and the range?

**Tip:** The range is maximized at a launch angle of 45°, but it varies if the launch angle or initial velocity changes.

A ball is launched at an angle of 45.0° above the horizontal. It reaches a maximum height of 40.0 m. What is the range of the projectile?

Learn how to solve projectile motion problems where a ball is launched at 45° and reaches a maximum height of 40 m. Discover the range of the projectile using kinematic equations and basic trigonometry. Ideal for Grades 10-12 students.

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