Projectile motion with Vo = 10 m/s, θ = 60°, and g = 10 m/s² yields parametric equations x = 5t and y = 5√3t - 5t². Trajectory equation: y = √3x - 5x²/125. Max height H = 25√3/2 m. Range R = 50√3/3 m.
Projectile Motion Analysis
Given:
Initial velocity (Vo) = 10 m/s
Angle (θ) = 60°
Acceleration due to gravity (g) = 10 m/s²
1. Parametric Equations:
The horizontal (x) and vertical (y) components of the projectile's motion are given by:
x = Vo * cos(θ) * t
y = Vo * sin(θ) * t - 1/2 * g * t^2
where:
t is the time elapsed
Parametric equations obtained are:
x = 5 * t
y = 5√3 * t - 5t^2
2. Equation of Trajectory:
Eliminating t from the parametric equations, we obtain the equation of the trajectory:
y = tan(θ) * x - g * x^2 / (2 * Vo^2 * cos^2(θ))
y = √3 * x - 5x^2 / 125
3. Maximum Height:
The maximum height (H) occurs when the vertical velocity (Vy) is zero. Vy is given by:
Vy = Vo * sin(θ) - g * t
Setting Vy to zero and solving for t, we get:
t = Vo * sin(θ) / g
Substituting this value of t in the equation for y, we get the maximum height:
H = Vo^2 * sin^2(θ) / (2 * g)
H = 25√3 / 2 m
4. Range:
The range (R) is the horizontal distance traveled by the projectile when it hits the ground. This occurs when y = 0. Solving the equation for y for t:
t = (2 * Vo * sin(θ)) / g
Substituting this value of t in the equation for x, we get the range:
R = Vo^2 * sin(2θ) / g
R = 50√3 / 3 m