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