Consider trajectories of projectiles on level ground in the presence of air resistance. For a fixed initial speed, the maximum range is obtained with the initial angle of about Ο 45$^o$. Ο 38$^o$. Ο 75$^o$. Ο 25$^o$.
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For launch speed v and angle θ the range is R = v^2 sin(2θ)/g, which is maximized when sin(2θ)=1, i.e. θ = 45°. Show more…
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Refer to the following figure and formulas.The figure shows the trajectory of an object projected upward from ground level at an angle $\alpha .$ Neglecting air resistance, the trajectory is approximately a parabola. The maximum height $h_{\max }$ and the range $r$ are given by the formulas adjacent to the figure. In the formulas, $v_{0}$ is the initial velocity; the constant $g$ is the acceleration due to gravity at the earth's surface. If distance is measured in feet, the value for $g$ is 32 Itsec^{2} { ; if distance is } measured in meters, the value for $g$ is $9.8 \mathrm{m} / \mathrm{sec}^{2}$/. $$\begin{aligned} h_{\max } &=\frac{v_{0}^{2} \sin ^{2} \alpha}{2 g} \\ r &=\frac{v_{0}^{2} \sin 2 \alpha}{g} \end{aligned}$$ (FIGURE CANNOT COPY) As background for this exercise, you need to have worked Exercise $89 .$ Assume that the initial velocity is $25 \mathrm{m} / \mathrm{sec} .$ Find the two angles that yield a range of $60 \mathrm{m}$ Express the answers in degrees, rounded to one decimal place.
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