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.