Question

A projectile is launched from the ground at an angle of theta above the horizontal with an initial speed of v in​ ft/s. The range​ (the total distance traveled by the projectile over level​ ground) of the projectile is approximated by the equation $x = \frac{32}{v^2} sin 2 \thetax= \ \textless \ br /\ \textgreater \ sin2θ$. Find the launch angle of a projectile with an initial speed of 85​ft/s and a range of 170 ft.​

Answer 1

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$\theta=24.4^circθ$

Further Explanation

The formula you wrote has the fraction the other way around. You can find that the range of a projectile is in reality approximated by the equation $x=\frac{v^2}{g} sin(2\theta)=\frac{v^2}{32ft/s^2} sin(2\theta)x=gv2sin(2θ)=32ft/s2v2sin(2θ)$, where we will use ft for distances.

From the given equation we have then $\frac{x32ft/s^2}{v^2}$ , which means $\theta=\frac{Arcsin(\frac{x32ft/s^2}{v^2})}{2}$since the arcsin is the inverse function of the sin.

Since we have x = 170 ft and  v = 85 ft/s, we can substitute these values from the equation written, and we will have $\theta=\frac{Arcsin(\frac{(170ft)(32ft/s^2)}{(85ft/s)^2})}{2}$ , and from now on we have just to use a calculator, obtaining $\theta=\frac{Arcsin(0.75294117647)}{2}=\frac{48.84579804^\circ}{2}=24.4^\circθ$

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