Projectile motion involves two-dimensional movement, which can be visualized as a trajectory that goes up and then down, resembling a parabola.
Understanding this motion requires breaking down the velocity into its components, the horizontal (x-direction) and vertical (y-direction).
A velocity vector (V) at an angle θ can be split into:
Horizontal Component (Vx): Vx = V cos(θ)
Vertical Component (Vy): Vy = V sin(θ)
Vx represents constant speed (no acceleration) in horizontal motion.
Vy is affected by gravity, leading to acceleration in vertical motion (downward).
Horizontal Motion: Constant velocity, so: Vx = displacement (Sx) / time (T).
Vertical Motion:
Uses equations of accelerated motion.
At maximum height, vertical velocity (Vy) becomes 0, while horizontal velocity (Vx) remains constant.
Scenario: Object launched from a 100m high cliff at 50 m/s at a 30° angle.
Break down initial velocity:
Vx = 50 cos(30°) = 43.3 m/s
Vy = 50 sin(30°) = 25 m/s
Maximum Height Calculation:
At max height, vertical speed is 0.
Using V² = U² + 2aS, where V=0, U=25 (initial vertical speed), a=-9.81 (gravity), solve for S:
S = -(U²)/(2a) = -(25²)/(2 * -9.81) = 31.855m
Total maximum height from ground = cliff height (100m) + vertical height (32m) = 132m.
Find horizontal displacement (Sx) during flight:
Use Sx = Vx * T, where T is the total time in the air.
Vertical Motion: Final displacement is 100m down:
S = Ut + ½at² yields:
100 = 25T + 4.95T², leading to a quadratic equation to solve for T.
After calculating, T ≈ 7.7 seconds.
Final horizontal distance: Sx = Vx * T = 43.3 m/s * 7.7 s = 333m (or 330m rounded).
Understanding the separation of components in projectile motion allows for methodical calculations of maximum height and horizontal distance.