suppose a projectile is fired from a cannon with velocity ( v_0 ) and angle of elevation ( \theta ). the…

suppose a projectile is fired from a cannon with velocity ( v_0 ) and angle of elevation ( \theta ). the horizontal distance ( r(\theta) ) it travels (in feet) is given by the following.\n\n r(\theta)=\frac{left(v_0\right)^2 sin 2 \theta}{32} \n\nif ( v_0 = 160 mathrm{ft} / mathrm{s} ), what angle ( \theta ) (in radians) should be used to hit a target on the ground 564 feet in front of the cannon?\n\ndo not round any intermediate computations, and round your answer(s) to the nearest hundredth of a radian. (if there is more than one answer, additional answers with the \or\ button.)\n\n \theta=square mathrm{rad}
Answer
Explanation:
Step1: Substitute the given values into the formula
Given (v_0 = 160) ft/s and (R(\theta)=564) feet. Substitute into (R(\theta)=\frac{(v_0)^2\sin2\theta}{32}). So, (564=\frac{(160)^2\sin2\theta}{32}). First, simplify (\frac{(160)^2}{32}): (\frac{25600}{32}=800). The equation becomes (564 = 800\sin2\theta).
Step2: Solve for (\sin2\theta)
Divide both sides of the equation (564 = 800\sin2\theta) by (800). (\sin2\theta=\frac{564}{800}=\frac{141}{200}=0.705).
Step3: Solve for (2\theta)
Since (\sin2\theta = 0.705), then (2\theta=\sin^{- 1}(0.705)) or (2\theta=\pi-\sin^{-1}(0.705)) (because (\sin x=\sin(\pi - x))). Using a calculator, (\sin^{-1}(0.705)\approx0.78) radians and (\pi-\sin^{-1}(0.705)\approx3.14 - 0.78=2.36) radians.
Step4: Solve for (\theta)
If (2\theta = 0.78) radians, then (\theta=\frac{0.78}{2}=0.39) radians. If (2\theta = 2.36) radians, then (\theta=\frac{2.36}{2}=1.18) radians.
Answer:
(\theta = 0.39) rad or (\theta=1.18) rad