$\text { The range } R \text { of a projectile is } R = \frac { v _ { 0 } ^ { 2 } } { 32 } ( \sin 2 \theta ) \text { where } v _ { 0 } \text { is the initial velocity in feet }$ per second and $\theta$ is the angle of elevation. If $v _ { 0 } = 2300$ feet per second and $\theta$ is changed from $13 ^ { \circ }$ to $14 ^ { \circ }$ use differentials to approximate the change in the range. Round your answer to the nearest integer.
A) $2,593 \mathrm { ft }$
B) $4,568 \mathrm { ft }$
C) $5,623 \mathrm { ft }$
D) $2,811 \mathrm { ft }$
E) $5,186 \mathrm { ft }$

Question

Quizplus · Accepted Answer

The differential $dR$ is given by $dR = \frac{v_0^2}{32} \cdot 2 \cos(2\theta) \cdot d\theta$. Substituting $v_0 = 2300$, $\theta = 13^\circ$, and $d\theta = 1^\circ = \frac{\pi}{180}$ radians, we find $dR \approx 5186$ ft.

Per Second And θ\thetaθ Is the Angle of Elevation v0=2300v _ { 0 } = 2300v0​=2300

Per Second And $\theta$ Is the Angle of Elevation $v _ { 0 } = 2300$