If you throw a stone into the air at an angle of $\theta$ to the horizontal, it moves along the curve $y=x \tan \theta-\frac{x^{2}}{100}\left(1+\tan ^{2} \theta\right)$ , where y is the height of the stone above the ground, x is the horizontal distance.If the angle $\theta$ is fixed, what value of x gives the maximum height? (Your answer will $\theta$.)
A) $\frac{50 \tan \theta}{1+\tan ^{2} \theta}$
B) $\frac{1+\tan ^{2} \theta}{50 \tan \theta}$
C) $\frac{\tan \theta}{50+50 \tan ^{2} \theta}$
D) $\frac{50+50 \tan ^{2} \theta}{\tan \theta}$

Question

Quizplus · Accepted Answer

To find the value of $x$ that gives the maximum height, we differentiate the given equation with respect to $x$ and set the derivative equal to zero. The derivative of $y$ with respect to $x$ is $\tan \theta - \frac{2x}{100}(1+\tan^2 \theta)$. Setting this equal to zero and solving for $x$ gives $x = \frac{50 \tan \theta}{1+\tan^2 \theta}$, which is option A.

If You Throw a Stone into the Air at an Angle