Question

15.) the confederates had at least one 24 lb mortar at the battle of gettysburg in the civil war. if the muzzle velocity of a projectile was 400 ft/sec, then at what angles, rounded to the nearest tenth, could the cannon be aimed to hit the union army 3000 ft away? hint: the distance d (in feet) traveled by a projectile fired at an angle θ is related to the initial velocity v₀ (in feet per second) by the equation ( v_0^2 sin(2\theta) = 32d ).

Explanation:

Step1: Substitute values into the formula

We know \( v_0 = 400 \) ft/sec and \( d = 3000 \) ft. Substitute these into the equation \( v_0^2\sin(2\theta)=32d \).
So we get \( 400^2\sin(2\theta)=32\times3000 \).
Calculate \( 400^2 = 160000 \) and \( 32\times3000 = 96000 \).
The equation becomes \( 160000\sin(2\theta)=96000 \).

Step2: Solve for \(\sin(2\theta)\)

Divide both sides of the equation \( 160000\sin(2\theta)=96000 \) by \( 160000 \).
\(\sin(2\theta)=\frac{96000}{160000}=\frac{3}{5} = 0.6\).

Step3: Find \(2\theta\)

We know that if \(\sin(x)=0.6\), then \(x=\arcsin(0.6)\) or \(x = 180^{\circ}-\arcsin(0.6)\) (in degrees) because sine is positive in the first and second quadrants.
First, calculate \(\arcsin(0.6)\approx36.87^{\circ}\).
So \(2\theta\approx36.87^{\circ}\) or \(2\theta\approx180^{\circ}- 36.87^{\circ}=143.13^{\circ}\).

Step4: Solve for \(\theta\)

For \(2\theta\approx36.87^{\circ}\), divide by 2: \(\theta\approx\frac{36.87^{\circ}}{2}\approx18.4^{\circ}\).
For \(2\theta\approx143.13^{\circ}\), divide by 2: \(\theta\approx\frac{143.13^{\circ}}{2}\approx71.6^{\circ}\).

Answer:

The angles are approximately \(18.4^{\circ}\) and \(71.6^{\circ}\)