Q:

# With what inclination must a body be thrown so that its horizontal range is equal to four times its maximum height?

Accepted Solution

A:
The horizontal range (R) of a projectile is given by the formula: R = (v^2 * sin(2θ)) / g where: v is the initial velocity of the projectile θ is the angle of inclination (launch angle) in radians g is the acceleration due to gravity The maximum height (H) reached by the projectile is given by the formula: H = (v^2 * sin^2(θ)) / (2 * g) Given that the horizontal range is four times the maximum height, we can write: R = 4 * H Substitute the expressions for R and H: (v^2 * sin(2θ)) / g = 4 * ((v^2 * sin^2(θ)) / (2 * g)) Simplify: sin(2θ) = 2 * sin^2(θ) Now, use the trigonometric identity for double angle: sin(2θ) = 2 * sin(θ) * cos(θ) Substitute this back into the equation: 2 * sin(θ) * cos(θ) = 2 * sin^2(θ) Divide both sides by 2 * sin(θ): cos(θ) = sin(θ) Now, using the trigonometric identity for complementary angles (cos(θ) = sin(90° - θ)), we have: sin(90° - θ) = sin(θ) This implies: 90° - θ = θ Solve for θ: 2θ = 90° θ = 45° Therefore, the body must be thrown at an angle of 45 degrees with the horizontal to achieve a horizontal range equal to four.