Solve the problem.
-The altitude of a projectile in feet (neglecting air resistance) is given by
$$y = ( \tan \theta ) x - \frac { 16 } { v ^ { 2 } \cos ^ { 2 } \theta } x ^ { 2 }$$
where $x$ is the horizontal distance covered in feet and $v$ is the initial velocity of the projectile at an angle $\theta$ from $t$ ] horizontal. Find the firing angle (in degrees) of a projectile fired at an initial velocity of 100 feet per second so that it strikes the ground $312.5$ feet from the firing point.
A) $30 ^ { \circ }$
B) $45 ^ { \circ }$
C) $50 ^ { \circ }$
D) $22.5 ^ { \circ }$

Question

Quizplus · Accepted Answer

The altitude $y$ becomes 0 when the projectile strikes the ground. Substituting $x = 312.5$ and $v = 100$ into the given equation and setting $y = 0$, we solve for $\theta$. The equation simplifies to a quadratic in terms of $\tan(\theta)$ and $\cos^2(\theta)$, which, when solved, gives the angle $\theta$ that satisfies the given conditions. The angle $45^\circ$ is known to maximize the range of a projectile under ideal conditions (without air resistance), which matches the conditions given in the problem.

Solve the Problem y=(tan⁡θ)x−16v2cos⁡2θx2y = ( \tan \theta ) x - \frac { 16 } { v ^ { 2 } \cos ^ { 2 } \theta } x ^ { 2 }y=(tanθ)x−v2cos2θ16​x2

Solve the Problem $y = ( \tan \theta ) x - \frac { 16 } { v ^ { 2 } \cos ^ { 2 } \theta } x ^ { 2 }$