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

Question 42

Multiple Choice

 The range R of a projectile is R=v0232(sin2θ)  where v0 is the initial velocity in feet \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 θ\theta is the angle of elevation. If v0=2300v _ { 0 } = 2300 feet per second and θ\theta is changed from 1313 ^ { \circ } to 1414 ^ { \circ } use differentials to approximate the change in the range. Round your answer to the nearest integer.


A) 2,593ft2,593 \mathrm { ft }
B) 4,568ft4,568 \mathrm { ft }
C) 5,623ft5,623 \mathrm { ft }
D) 2,811ft2,811 \mathrm { ft }
E) 5,186ft5,186 \mathrm { ft }

Correct Answer:

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