A certain projectile is launched with an initial speed Vo. At its highest point its speed is Vo / 6.
What was the launch angle?
To determine the launch angle of the projectile, we can use the concept of projectile motion and the conservation of energy.
At the highest point of the projectile's path, the speed is given as Vo/6. This speed occurs when the vertical velocity of the projectile becomes zero. Using this information, we can set up the following equation:
Vo^2/2 - g * h = (Vo/6)^2/2
Where Vo is the initial speed of the projectile, g is the acceleration due to gravity, and h is the maximum height reached by the projectile.
Since the projectile reaches its highest point, the vertical displacement (h) can be determined using the equation for the maximum height of a projectile:
h = (Vo^2 * sin^2(theta)) / (2 * g)
Where theta is the launch angle.
By substituting this value of h into our previous equation, we get:
Vo^2/2 - g * [(Vo^2 * sin^2(theta)) / (2 * g)] = (Vo/6)^2/2
Simplifying the equation, we have:
Vo^2 * (1 - sin^2(theta)) = Vo^2/36
Cancelling out the Vo^2 terms, we get:
1 - sin^2(theta) = 1/36
Rearranging the equation, we have:
sin^2(theta) = 1 - 1/36
sin^2(theta) = 35/36
Taking the square root of both sides, we have:
sin(theta) = sqrt(35/36)
Now, to find the launch angle (theta), we can take the inverse sine (sin^(-1)) of both sides:
theta = sin^(-1)(sqrt(35/36))
Using a calculator, we can find the inverse sine of sqrt(35/36) to determine the launch angle.