Solve the problem.
-The path of a projectile fired at an inclination $\theta$ to the horizontal with an initial speed $v _ { O }$ is a parabola. The range $\mathrm { R }$ of the projectile, the horizontal distance that the projectile travels, is found by the formula $\mathrm { R } = \frac { \mathrm { v } _ { \mathrm { O } } ^ { 2 } \sin 2 \theta } { \mathrm { g } }$ where $\mathrm { g } = 32.2$ feet per second per second or $\mathrm { g } = 9.8$ meters per second per second. Find the range of a projectile fired with an initial velocity of 140 feet per second at an angle of $27 ^ { \circ }$ to the horizontal. Round your answer to two decimal places.
A) $276.34 \mathrm { ft }$
B) $492.35 \mathrm { ft }$
C) $492.45 \mathrm { ft }$
D) $552.68 \mathrm { ft }$

Question

Quizplus · Accepted Answer

To find the range, use the formula $ R = \frac{v_0^2 \sin 2\theta}{g} $. Here, $ v_0 = 140 $ ft/s, $ \theta = 27^\circ $, and $ g = 32.2 $ ft/s². First, calculate $ 2\theta = 54^\circ $. Then, find $ \sin 54^\circ \approx 0.809 $. Substitute into the formula: $ R = \frac{140^2 \times 0.809}{32.2} \approx 492.45 $ ft.

Solve the Problem θ\thetaθ To the Horizontal with an Initial Speed vOv _ { O }vO​

Solve the Problem $\theta$ To the Horizontal with an Initial Speed $v _ { O }$