The projectile motion is fired with velocity of magnitude vo at the angle feta. Find feta for which the maximum elevation of the projectile is twice its range

Question

The projectile motion is fired with velocity of magnitude vo at the angle feta. Find feta for which the maximum elevation of the projectile is twice its range

Answer 1

feta is a cheese, not much else. Theta is a Greek letter that is often used for an angle. But feta? it won't hold up more than about 30 degrees, and even then if falls down.

Vertical height=2*horiztal distance.

So vertical height first:

at the top, vertical velocity is zero.
vtop=vo*sinTheta-g t
or time to top is vo*sinTheta/g

then verticalmax height is
hmax=vo*sinTheta*vo*sinTheta/g

horizontal distance:
vo*cosTheta* vo sinTheta/g

then
vertical max=2*horizontal
(vo*sinTheta)^2 /g = 2*(vo^2 cosTheta*sinTheta)/g

or sintheta=2costheta

tan Theta=2
theta=arcTan2 or = 63.4349488 degrees

check all this.

Answer 2

To find the angle feta for which the maximum elevation of the projectile is twice its range, we can use the equations of projectile motion.

Let's denote the maximum elevation as h and the range as R.

First, let's find the equations for maximum elevation and range in terms of the given parameters.

The maximum elevation can be determined using the equation:
h = (vo^2 * sin^2(feta)) / (2 * g)

The range can be determined using the equation:
R = (vo^2 * sin(2 * feta)) / g

Now, we need to find the angle feta for which the maximum elevation is twice the range. Mathematically, this can be expressed as:

2 * R = h

Substituting the equations for h and R, we get:

2 * (vo^2 * sin(2 * feta)) / g = (vo^2 * sin^2(feta)) / (2 * g)

Simplifying the equation:

2 * (sin(2 * feta)) = sin^2(feta)

Expanding the left side of the equation:

2 * (2 * sin(feta) * cos(feta)) = sin^2(feta)

Simplifying the equation further:

4 * sin(feta) * cos(feta) = sin^2(feta)

Dividing both sides of the equation by sin(feta) (assuming sin(feta) ≠ 0):

4 * cos(feta) = sin(feta)

Rearranging the equation:

sin(feta) = 4 * cos(feta)

Dividing both sides of the equation by cos(feta) (assuming cos(feta) ≠ 0):

tan(feta) = 4

To find feta, we can take the inverse tangent (arctan) of both sides:

feta = arctan(4)

Using a calculator or mathematical software, we can find the value of arctan(4) to be approximately 1.3258 radians or 75.96 degrees.

Therefore, the angle feta for which the maximum elevation of the projectile is twice its range is approximately 1.3258 radians or 75.96 degrees.