11. The horizontal range of a projectile is 4 3 times of its maximum height. Calculate

the angle of projection.

To calculate the angle of projection, we can use the equations for horizontal range and maximum height of a projectile.

Let's denote the angle of projection as theta (θ).
The horizontal range (R) of a projectile is given by:
R = (u^2 * sin(2θ)) / g -------------- (Equation 1)

The maximum height (H) of a projectile is given by:
H = (u^2 * sin^2(θ)) / (2g) -------------- (Equation 2)

Given that the horizontal range is 4 3 times the maximum height, we can write:
4 3H = R

Therefore, from Equations 1 and 2, we have:
4 3 * ((u^2 * sin^2(θ)) / (2g)) = (u^2 * sin(2θ)) / g

Canceling out u^2 and g from both sides, we get:
4 3 * sin^2(θ) = sin(2θ)

Expanding sin(2θ) using the double-angle formula sin(2θ) = 2sin(θ)cos(θ), we get:
4 3 * sin^2(θ) = 2sin(θ)cos(θ)

Dividing both sides by 2sin(θ), we get:
4 3 * sin(θ) = 2cos(θ)

Dividing both sides by cos(θ), we get:
4 3 * tan(θ) = 2

Dividing both sides by 4 3, we get:
tan(θ) = 2/4 3 = 6/4 = 3/2

To find the angle of projection, we can take the inverse tangent (arctan) of both sides:
θ = arctan(3/2)

Using a calculator, we find that the angle of projection (θ) is approximately 56.31 degrees.

Therefore, the angle of projection is approximately 56.31 degrees.

To answer this question, we will use the formulas for the horizontal range and the maximum height of a projectile.

Let's assume the angle of projection is represented by θ. The horizontal range, denoted by R, is given by:

R = (v^2 * sin(2θ)) / g

where v is the initial velocity of the projectile and g is the acceleration due to gravity.

The maximum height, denoted by H, is given by:

H = (v^2 * sin^2(θ)) / (2g)

Now, given that the horizontal range is 4 3 times the maximum height, we can write the equation as:

4 3 * H = R

Substituting the formulas for R and H, we get:

4 3 * ((v^2 * sin^2(θ)) / (2g)) = (v^2 * sin(2θ)) / g

Next, we can simplify the equation by canceling out common terms:

4 3 * sin^2(θ) / 2 = sin(2θ)

To solve for θ, we need to eliminate the sin(2θ) term. We can use the double-angle identity for sine:

sin(2θ) = 2*sin(θ)*cos(θ)

Substituting this back into the equation, we have:

4 3 * sin^2(θ) / 2 = 2*sin(θ)*cos(θ)

Now, we can cancel out the sin(θ) term:

4 3 * sin(θ) / 2 = 2*cos(θ)

Multiply both sides of the equation by 2/3 to isolate sin(θ):

2/3 * 4 3 * sin(θ) / 2 = 2/3 * 2*cos(θ)

4 3 * sin(θ) / 3 = 4/3 * cos(θ)

Divide both sides by 4/3 to isolate θ:

sin(θ) / 3 = cos(θ)

Divide both sides by cos(θ):

tan(θ) = 3

Now, we need to find the angle whose tangent is 3. Using a calculator, we can take the inverse tangent (arctan) of 3:

θ = arctan(3)

Calculating this gives us the value of the angle of projection, θ.

Note: Make sure to set your calculator to the appropriate angle mode (degrees or radians) based on the problem requirements.