Answer:
0.25 rad to the nearest hundredth radian
Step-by-step explanation:
Here is the complete question
Suppose a projectile is fired from a cannon with velocity vo and angle of elevation (theta). The horizontal distance R(θ) it travels (in feet) is given by the following.
R(θ) = v₀²sin2θ/32
If vo=80ft/s what angel (theta) (in radians) should be used to hit a target on the ground 95 feet in front of the cannon?
Do not round any intermediate computations, and round your answer(s) to the nearest hundredth of a radian.
(θ)= ?rad
Solution
R(θ) = v₀²sin2θ/32
If v₀ = 80 ft/s and R(θ) = 95 ft
θ = [sin⁻¹(32R(θ)/v₀²)]/2
= [sin⁻¹(32 × 95/80²)]/2
= [sin⁻¹(3040/6400)]/2
= [sin⁻¹(0.475)]/2
= 28.36°/2
= 14.18°
Converting 14.18° to radians, we have 14.18° × π/180° = 0.2475 rad
= 0.25 rad to the nearest hundredth radian
Answer:
Step-by-step explanation:
the second one , why? because 7a-12a=-5a
4 - 8 = -4
Answer:
C. ED
Step-by-step explanation:
The center of the angle is the point E, so it would be written first.
Answer:
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