MATH SOLVE

5 months ago

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.