A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is
40-m/s at an angle of 37º with respect to the horizontal and the cannonball is 7.0-m above the ground at
this time.
a) What is the maximum height attained by the cannonball?
b) Assuming the cannon ball clears the wall of the castle, what is the horizontal distance this cannonball
was shot?

We will approach this problem using projectile motion equations, breaking the motion into horizontal and vertical components. The key data provided is:

- Initial velocity (\(v_0\)) = 40 m/s
- Angle of projection (\(\theta\)) = 37º
- Initial height (\(y_0\)) = 7.0 m
- Gravitational acceleration (\(g\)) = 9.8 m/s²

### Part (a): Maximum height attained by the cannonball

First, we need to find the vertical component of the initial velocity, which is given by:
\[
v_{0y} = v_0 \sin(\theta)
\]
Substituting the values:
\[
v_{0y} = 40 \sin(37^\circ) \approx 40 \times 0.6018 = 24.07 \, \text{m/s}
\]

The formula to calculate the maximum height is:
\[
h_{\text{max}} = y_0 + \frac{v_{0y}^2}{2g}
\]
Where:
- \(y_0 = 7.0 \, \text{m}\) (initial height)
- \(v_{0y} = 24.07 \, \text{m/s}\) (vertical component of velocity)
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)

Now, calculate the maximum height:
\[
h_{\text{max}} = 7.0 + \frac{(24.07)^2}{2 \times 9.8}
\]
\[
h_{\text{max}} = 7.0 + \frac{579.36}{19.6}
\]
\[
h_{\text{max}} = 7.0 + 29.56 \approx 36.56 \, \text{m}
\]

So, the **maximum height** attained by the cannonball is approximately **36.56 meters**.

---

### Part (b): Horizontal distance (Range)

To find the horizontal distance the cannonball travels, we need to find the time of flight. The vertical displacement equation for projectile motion is:
\[
y = y_0 + v_{0y}t - \frac{1}{2}gt^2
\]
Since the cannonball lands back on the ground, \(y = 0\), so the equation becomes:
\[
0 = 7.0 + 24.07t - \frac{1}{2} \times 9.8 t^2
\]
Simplifying:
\[
0 = 7.0 + 24.07t - 4.9t^2
\]
This is a quadratic equation of the form:
\[
4.9t^2 - 24.07t - 7.0 = 0
\]
Solving this using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where:
- \(a = 4.9\)
- \(b = -24.07\)
- \(c = -7.0\)

Substitute the values:
\[
t = \frac{-(-24.07) \pm \sqrt{(-24.07)^2 - 4(4.9)(-7.0)}}{2(4.9)}
\]
\[
t = \frac{24.07 \pm \sqrt{579.36 + 137.2}}{9.8}
\]
\[
t = \frac{24.07 \pm \sqrt{716.56}}{9.8}
\]
\[
t = \frac{24.07 \pm 26.76}{9.8}
\]
Taking the positive root (since time cannot be negative):
\[
t = \frac{24.07 + 26.76}{9.8} \approx \frac{50.83}{9.8} \approx 5.19 \, \text{seconds}
\]

Now, we calculate the horizontal distance traveled. The horizontal component of the initial velocity is:
\[
v_{0x} = v_0 \cos(\theta)
\]
Substituting the values:
\[
v_{0x} = 40 \cos(37^\circ) \approx 40 \times 0.7986 = 31.94 \, \text{m/s}
\]

The horizontal distance (range) is then:
\[
R = v_{0x} \times t
\]
Substitute the values:
\[
R = 31.94 \times 5.19 \approx 165.77 \, \text{m}
\]

So, the **horizontal distance** the cannonball was shot is approximately **165.77 meters**.

---

### Summary:
- **(a)** Maximum height: 36.56 m
- **(b)** Horizontal distance: 165.77 m

Let me know if you need further details or have any questions!

#### Related Questions:
1. How would a different angle of projection affect the range?
2. How would air resistance alter the cannonball’s motion?
3. What would happen if the initial velocity was increased?
4. How would the motion change if the ground level was sloped?
5. What is the total time of flight for the cannonball?

**Tip:** The angle of 45º provides the maximum range for projectile motion if the launch and landing heights are equal.

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 40 m/s at an angle of 37º with respect to the horizontal and the cannonball is 7.0 m above the ground at this time.

a) What is the maximum height attained by the cannonball?
b) Assuming the cannonball clears the wall of the castle, what is the horizontal distance this cannonball was shot?

Learn how to solve a projectile motion problem involving a cannonball launched at 40 m/s and an angle of 37º from a height of 7 m. This problem covers key concepts in kinematics, including maximum height calculation and horizontal range estimation, using projectile motion equations. Suitable for students in Grades 9-12.

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