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.

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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.

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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.