Answer:
89.1° or -1.4°
Step-by-step explanation:
1. Location:
You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.
The French troops are about 50 m below you and 1.2 km distant.
2. Finding the firing angle
Data:
R = 1200 m
u = 600 m/s
h = -50 m (the height of the target)
a = 9.8 m/s²
We have two conditions.
Horizontal distance
(1) 1200 = 600t cosθ
Vertical distance
(2) -50 = 600t sinθ - 4.9t²
Divide each side of (1) by 600cosθ.
[tex](3) \, t =\dfrac{2}{\cos \theta}[/tex]
Substitute (3) into (2)
[tex]-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}[/tex]
Recall that
(5) sec²θ = 1/cos²θ = tan²θ + 1
Substitute (5) into (4)
[tex]-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )[/tex]
Set up a quadratic equation
[tex]\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}[/tex]
Solve for θ
Use the quadratic formula.
tanθ = 61.249 or -0.025
θ = arctan(61.249) = 89.1° or
θ = arctan(-0.025) = -1.4°
