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
Hi,
The answer would be unit C. The distance between -80 & 0 is 80. The distance from 0 to 15 is 15. So 80+15=95. Hope this helps.
<span>False.
Experimental probability is based on doing trials.</span>
Yes because 18% is an abbysmal amount for a tip
Y=0.5x-0.5
( u find the slope using rise/run then after finding the slope u use the equation and a point on the graph to find the y-intercept
