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 $x = \frac{32}{v^2} sin 2 \theta$. Find the launch angle of a projectile with an initial speed of 85​ft/s and a range of 170 ft.

Answer 1

Answer:

$\theta=24.4^\circ$

Step-by-step 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 $x=\frac{v^2}{g} sin(2\theta)=\frac{v^2}{32ft/s^2} sin(2\theta)$, where we will use ft for distances.

From the given equation we have then $\frac{x32ft/s^2}{v^2} =sin(2\theta)$, 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 $\theta=\frac{Arcsin(0.75294117647)}{2}=\frac{48.84579804^\circ}{2}=24.4^\circ$