Show that the projectile angele thita for a projectile launched from origin is given by thita= tan^ minus 1(4H/R). Whete H is the maximum height attained by the projectile and R is the range
the article in wikipedia gives the formulas, which I'm sure you already have. Just divide H/R and simplify.
https://en.wikipedia.org/wiki/Trajectory
To derive the equation for the projectile angle θ, we can analyze the motion of the projectile and make use of trigonometry.
Let's consider a projectile launched from the origin. The maximum height attained by the projectile is H, and the range (horizontal distance traveled) is R.
The motion of the projectile can be divided into two components: horizontal motion and vertical motion.
Horizontal Motion:
The time t taken for the projectile to reach the maximum height is the same as the time taken to fall back to the ground.
Since there is no acceleration in the horizontal direction, the horizontal velocity Vx remains constant throughout the motion.
The horizontal distance traveled is given by R = Vx * t.
From the equation for time in projectile motion, t = (2 * V0 * sin(θ)) / g, where V0 is the initial velocity of the projectile and g is the acceleration due to gravity.
Substituting this into the equation, we have R = V0 * cos(θ) * ((2 * V0 * sin(θ)) / g).
Vertical Motion:
The time t taken for the projectile to reach the maximum height is given by t = (V0 * sin(θ)) / g.
Since the only force acting vertically is gravity, the vertical displacement is given by H = ((V0 * sin(θ))^2) / (2 * g).
We can solve the second equation for V0 * sin(θ) and substitute it into the first equation:
V0 * sin(θ) = sqrt(2 * g * H)
Substituting this back into R = V0 * cos(θ) * ((2 * V0 * sin(θ)) / g), we have R = V0^2 * sin(2θ) / g.
Now we can solve for θ:
R = V0^2 * sin(2θ) / g
R * g = V0^2 * sin(2θ)
sin(2θ) = (R * g) / V0^2
2θ = sin^(-1)((R * g) / V0^2)
θ = sin^(-1)((R * g) / (2 * V0^2))
Using the relationship V0^2 = H * g (which can be derived from the second equation for vertical motion), we can simplify further:
θ = sin^(-1)((R * g) / (2 * V0^2))
θ = sin^(-1)((R * g) / (2 * (H * g)))
θ = sin^(-1)((R) / (2 * H))
θ = tan^(-1)((R) / (2 * H))
Therefore, the equation for the projectile angle θ, when a projectile is launched from the origin, is θ = tan^(-1)((R) / (2 * H)).