The problem is about projectile motion. In projectile motion, the horizontal range (R) and the maximum height (H) are given by the following formulas:

R = (v^2 * sin(2θ))/g
H = (v^2 * sin^2(θ))/2g

where:
- v is the initial velocity,
- θ is the angle of projection,
- g is the acceleration due to gravity.

The problem states that the horizontal range is equal to the maximum height, so we can set R = H and solve for θ:

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

Solving this equation gives:

sin(2θ) = 2 * sin^2(θ)

Using the double-angle formula, sin(2θ) = 2sin(θ)cos(θ), we can rewrite the equation as:

2sin(θ)cos(θ) = 2 * sin^2(θ)

Dividing both sides by 2sin(θ) gives:

cos(θ) = sin(θ)

This equation is true when θ = 45 degrees. Therefore, the angle of projection is 45 degrees.

Question

The problem is about projectile motion. In projectile motion, the horizontal range (R) and the maximum height (H) are given by the following formulas:

R = (v^2 * sin(2θ))/g
H = (v^2 * sin^2(θ))/2g

where:
- v is the initial velocity,
- θ is the angle of projection,
- g is the acceleration due to gravity.

The problem states that the horizontal range is equal to the maximum height, so we can set R = H and solve for θ:

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

Solving this equation gives:

sin(2θ) = 2 * sin^2(θ)

Using the double-angle formula, sin(2θ) = 2sin(θ)cos(θ), we can rewrite the equation as:

2sin(θ)cos(θ) = 2 * sin^2(θ)

Dividing both sides by 2sin(θ) gives:

cos(θ) = sin(θ)

This equation is true when θ = 45 degrees. Therefore, the angle of projection is 45 degrees.

Knowee AI · Accepted Answer