The horizontal distance d (in feet) traveled by a projectile with an initial speed of v feet per second is modeled by $d = \frac { v ^ { 2 } } { 32 } \sin 2 \theta$ , where $\theta$ is the angle at which the projectile is launched.
Find the horizontal distance traveled by a golf ball that is hit with an initial speed of 80 feet per second when the ball is hit at an angle of $\theta = 60 ^ { \circ }$ .Round to the nearest foot.
A)260
B)346
C)173
D)303
E)87

Question

Quizplus · Accepted Answer

To find the horizontal distance $d$, substitute $v = 80$ feet per second and $\theta = 60^\circ$ into the formula $d = \frac{v^2}{32} \sin 2\theta$. This gives $d = \frac{80^2}{32} \sin 120^\circ$. Since $\sin 120^\circ = \sin (180^\circ - 60^\circ) = \sin 60^\circ = \sqrt{3}/2$, the calculation simplifies to $d = \frac{6400}{32} \cdot \frac{\sqrt{3}}{2} = 200 \cdot \frac{\sqrt{3}}{2} = 100\sqrt{3}$. Approximating $\sqrt{3} \approx 1.732$, we get $d \approx 100 \cdot 1.732 = 173.2$, which rounds to 173 feet.

The Horizontal Distance D (In Feet) Traveled by a Projectile d=v232sin⁡2θd = \frac { v ^ { 2 } } { 32 } \sin 2 \thetad=32v2​sin2θ

The Horizontal Distance D (In Feet) Traveled by a Projectile $d = \frac { v ^ { 2 } } { 32 } \sin 2 \theta$