Suppose a projectile is fired from a cannon with velocity v0 and angle of elevation θ. The horizontal distance Rθ it travels (in feet) is given by the following. =Rθv02sin2θ32
If =v080 /fts, what angle θ (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. (If there is more than one answer, enter additional answers with the "or" button.)

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

the answer on ed. is:

c. "the graph of f(x) has exactly two x-intercepts."

hope i could someone! : )