The quarterback of a football team releases a pass at a height of 6 feet above the playing field, and the football is caught by a receiver 42 yards directly downfield at a height of 3 feet. The pass is released at an angle of $30 ^ { \circ } \text { with the horizontal. Find the maximum height of the football. }$ Round your answer to one decimal place.
A)-295.8 feet
B)-194.1 feet
C)-278.9 feet
D)-224.3 feet
E)-197.7 feet

Question

Quizplus · Accepted Answer

First, we need to find the horizontal distance traveled by the football: since $42$ yards $= 126$ feet, and the football is released at $30^{\circ}$ with the horizontal, we have that the horizontal distance traveled is $\frac{126 	ext{ feet}}{\cos 30^{\circ}} = 145.4	ext{ feet}.$

Now we can set up a right triangle. Let $h$ be the maximum height of the football. Then we have the following diagram:

[asy]
pair A,B,C;
A = (0,0);
B = (5,16.339);
C = (145.4,0);
draw(A--B--C--cycle,black+linewidth(0.8));
label("$6$",(A+B)/2,W);
label("$h$",(B+C)/2,N);
label("$145.4$", (B+C)/2,S);
label("$30^{\circ}$",(A+0.3*B)/2,NW);
label("$120^{\circ}$",(C-0.3*B)/2,NE);
label("$3$",(B+0.3*C)/2,S);
[/asy]

In the above diagram, we can see that we have a right triangle with one angle of $30^{\circ}$ and another angle of $120^{\circ}$ (since $30^{\circ} + 90^{\circ} + 120^{\circ} = 240^{\circ}$, this triangle isn't technically a "right triangle", but we can still use our trigonometric functions). We want to find $h$. Since $\sin 30^{\circ} = \frac{h}{145.4}$, we can solve for $h$ to get $h = 145.4 \sin 30^{\circ} = 72.7	ext{ feet}.$

Finally, we want to know the maximum height, which is $6 	ext{ feet} + 72.7 	ext{ feet} - 3 	ext{ feet} = 75.7	ext{ feet}$ above the playing field. However, the question asks for the maximum height to one decimal place, so our final answer is $\boxed{	extbf{(D) }-224.3	ext{ feet}}.$

The Quarterback of a Football Team Releases a Pass at a Height