Solve the problem.
-The optimal angle of elevation $\theta$ that a shot-putter should aim for in order to throw the greatest distance depends on the velocity $\mathrm { v }$ of the throw and the initial height $\mathrm { h }$ of the shot. One model for $\theta$ that achieves this maximum distance is $\theta = \sin ^ { - 1 } \left( \sqrt { \frac { \mathrm { v } ^ { 2 } } { 2 \mathrm { v } ^ { 2 } + 64 \mathrm {~h} } } \right)$. Suppose a shot-putter can consistently release the steel ball with velocity of 34 feet per second from an initial height $h$ of $6.1$ feet. What angle, to the nearest degree, will maximize the distance?

Question

Quizplus · Accepted Answer

To find the optimal angle $\theta$, substitute $v = 34$ feet per second and $h = 6.1$ feet into the given formula: $\theta = \sin ^ { - 1 } \left( \sqrt { \frac { 34 ^ { 2 } } { 2 \cdot 34 ^ { 2 } + 64 \cdot 6.1 } } \right)$. Simplifying inside the square root gives $\theta = \sin^{-1}\left(\sqrt{\frac{1156}{1156 + 390.4}}\right)$, which further simplifies to $\theta = \sin^{-1}\left(\sqrt{\frac{1156}{1546.4}}\right)$. Calculating the square root and then the arcsine gives an angle close to $41^\circ$, rounding to the nearest degree.

Solve the Problem θ\thetaθ That a Shot-Putter Should Aim for in Order to Throw

Solve the Problem $\theta$ That a Shot-Putter Should Aim for in Order to Throw