A projectile is launched up and to the right over flat, level ground. Its range is equal to half of its maximum elevation above the ground. What was the angle between its initial velocity and the ground? Ignore air resistance.


° above the horizonontal

max H = Vo^2 sin^2(Θ) / 2g

range = Vo^2 sin(2Θ) / g

2 Vo^2 sin(2Θ) / g = Vo^2 sin^2(Θ) / 2g

4 sin(2Θ) = sin^2(Θ)

solve for Θ

How can I solve for theta when there is no number A projectile is launched up and to the right over flat, level ground. Its range is equal to half of its maximum elevation above the ground. What was the angle between its initial velocity and the ground? Ignore air resistance.


° above the horizonontal

To find the angle between the initial velocity and the ground, we can use the concept of projectile motion and the equations of motion.

Let's consider the horizontal and vertical components of the projectile's motion independently.

The horizontal component of the initial velocity remains constant throughout the motion as there is no acceleration in that direction. Let's call it V0x.

The vertical component of the initial velocity, V0y, will change due to the acceleration due to gravity. At the highest point of the projectile's trajectory, V0y will be reduced to zero.

Now, let's define the variables:
θ = angle between the initial velocity and the ground
Range = R (given as half the maximum elevation)
Maximum Elevation = H

The range (R) of the projectile is given as half of its maximum elevation (H), so we can say: R = H/2.

We know that the horizontal range of a projectile can be calculated using the equation:
R = V0x * t, where t is the total time of flight.

To simplify our calculation, we can find the time of flight in terms of the initial vertical velocity (V0y) using the formula:
t = (2 * V0y) / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the equation for time into the range equation, we have:
R = V0x * ((2 * V0y) / g)

Since we are looking for the angle θ, we need to express V0x and V0y in terms of θ:
V0x = V0 * cos(θ)
V0y = V0 * sin(θ)

Substituting these expressions into the range equation, we get:
R = (V0 * cos(θ)) * ((2 * V0 * sin(θ)) / g)

Now, we can solve this equation for θ.
Dividing both sides by V0, we have:
R / V0 = (2 * sin(θ) * cos(θ)) / g

Using the identity sin(2θ) = 2 * sin(θ) * cos(θ), we can rewrite the equation as:
R / V0 = sin(2θ) / g

Finally, to solve for θ, we take the inverse sine (or arcsine) of both sides:
sin^(-1)((R / V0) * g) = 2θ

To find θ, divide the result by 2:
θ = (1/2) * sin^(-1)((R / V0) * g)

Now, you can calculate the angle using the given values for R (Range) and V0 (initial velocity). Remember to convert the angle from radians to degrees.

Keep in mind that this calculation assumes there is no air resistance and that the ground is flat and level.