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 θ from the 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°
B) 45°
C) 50°
D) 22.5°

Question

Quizplus · Accepted Answer

The projectile strikes the ground when $y = 0$. Substituting $x = 312.5$ feet and $v = 100$ feet per second into the given equation, we get $0 = (\tan \theta) \cdot 312.5 - \frac{16}{100^2 \cos^2 \theta} \cdot 312.5^2$. Solving this equation for $\theta$ requires finding the angle that satisfies the equation, which is $\theta = 45^\circ$, because at $45^\circ$, the range of a projectile is maximized for a given velocity, and the given equation simplifies due to the trigonometric identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, which reaches its maximum value when $\theta = 45^\circ$.

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