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Solve the Problem r=132v2sin2θ,\mathrm { r } = \frac { 1 } { 32 } \mathrm { v } ^ { 2 } \sin 2 \theta ,

Question 34

Multiple Choice

Solve the problem.
-The range r of a projectile is given by r=132v2sin2θ,\mathrm { r } = \frac { 1 } { 32 } \mathrm { v } ^ { 2 } \sin 2 \theta ,
where v\mathrm { v } is the initial velocity and θ\theta is the angle of elevation. If rr is to be 3000ft3000 \mathrm { ft } and v=2000ft/sec\mathrm { v } = 2000 \mathrm { ft } / \mathrm { sec } , what must the angle of elevation be? Give your answer in degrees to the nearest hundredth.


A) 89.3189.31 ^ { \circ }
B) 1.381.38 ^ { \circ }
C) 0.960.96 ^ { \circ }
D) 0.690.69 ^ { \circ }

Correct Answer:

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