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 = (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

`\theta=24.4^\circ`

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=(v^2)/(g) sin(2\theta)=(v^2)/(32ft/s^2) sin(2\theta)`, where we will use ft for distances.

From the given equation we have then
`(x32ft/s^2)/(v^2) =sin(2\theta)`, which means
`\theta=(Arcsin((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=(Arcsin(((170ft)(32ft/s^2))/((85ft/s)^2)))/(2)`, and from now on we have just to use a calculator, obtaining
`\theta=(Arcsin(0.75294117647))/(2)=(48.84579804^\circ)/(2)=24.4^\circ`