Q.Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola.

To fully determine a particle's trajectory (path in space), you need all three pieces of initial-condition information together: where it starts, how fast/which way it starts moving, and what acceleration acts on it throughout.

Given a constant acceleration a, the position at time t is found from:

r(t) = r0 + v0 t + (1/2) a t^2

• Initial position r0 fixes WHERE the path is located in space (translates the whole curve).
• Initial velocity v0 fixes the direction the object starts moving in, which combines with the acceleration direction to decide whether the path is a straight line (v0 parallel/antiparallel to a) or curved, e.g. a parabola (v0 not parallel to a). …

Horizontal launch = horizontal velocity component constant, vertical motion accelerates under gravity -- combining a linear (x) relation with a quadratic (y) relation in time always produces a parabola.

Let the object be launched horizontally with speed u from height h.

• Horizontal: x = u t (no horizontal acceleration)
• Vertical: y = (1/2) g t^2 (falling under gravity, starting with zero vertical velocity) …

At the highest point of a projectile's path, the vertical velocity component is momentarily zero, leaving only the horizontal component -- while gravity's acceleration always points straight down, so the two vectors are at 90 degrees.

For a projectile launched at an angle, the vertical component of velocity (v_y) decreases due to gravity, becomes zero exactly at the top of the trajectory, and then reverses sign on the way down.

At the topmost point: v_y = 0, so the velocity vector is entirely horizontal (v = v_x, in the horizontal direction).

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At the top of a projectile's trajectory, the vertical velocity is momentarily zero (that is why it's the highest point) — only the constant horizontal component $u\cos\theta$ remains.

A projectile launched with speed $u$ at angle $\theta$ to the horizontal has two independent components of velocity:

• Horizontal: $u_x = u\cos\theta$ (constant throughout the flight, since there is no horizontal force under gravity alone)
• Vertical: $u_y = u\sin\theta$, which decreases under gravity as $v_y = u\sin\theta - gt$ …

The range of a projectile is maximum at a projection angle of 45°.

The horizontal range of a projectile launched with speed u at angle θ to the horizontal is:

$R = \dfrac{u^2 \sin(2\theta)}{g}$

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The upward (vertical) speed of a projectile keeps decreasing due to gravity and becomes zero exactly at the highest point of the trajectory.

For a projectile launched with initial speed u at angle theta to the horizontal, the vertical component of velocity at time t is:

v_y = u sin(theta) - g t

At the start (t = 0), v_y = u sin(theta), which is the maximum upward speed.

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At maximum height the vertical velocity is zero and only the horizontal component u cos(theta) remains.

The initial velocity u makes angle theta with the horizontal, so its components are $u\cos\theta$ (horizontal) and $u\sin\theta$ (vertical). …

Gravity only acts vertically (downward), so it never changes the horizontal velocity component — it stays constant throughout the projectile's flight.

In projectile motion (ignoring air resistance), the only force acting is gravity, which is purely vertical (g, downward). Newton's second law tells us acceleration only exists along the direction of net force, so: …

The only force on a projectile is gravity, acting vertically downward. With zero horizontal force, the horizontal velocity component never changes -- everything else (KE, momentum, vertical velocity) changes as the projectile rises and falls.

In projectile motion, the acceleration due to gravity, g, acts purely in the vertical direction. There is no horizontal acceleration.

• Horizontal component of velocity, v_x = u cos(theta): remains CONSTANT throughout the flight, since a_x = 0. …

At the top of the trajectory the vertical velocity component is momentarily zero, leaving only the horizontal velocity component -- which is perpendicular to the ever-present, purely vertical, gravitational acceleration.

At the highest point of a projectile's path, the vertical component of velocity, v_y = u sin(theta) - gt, has dropped to zero (the projectile is momentarily neither rising nor falling). What remains is purely the horizontal component, v_x = u cos(theta), so the velocity vector at that instant points horizontally.

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Maximum height H = u^2 sin^2(theta) / (2g) is largest when sin(theta) = 1, i.e. theta = 90 degrees.

For a projectile launched with speed u at angle theta to the horizontal, the maximum height reached is:

H = u^2 sin^2(theta) / (2g)

For a fixed launch speed u, H depends only on sin^2(theta). This is maximum when sin(theta) = 1, which occurs at theta = 90 degrees. At this angle the projectile has no horizontal component of velocity at all -- it is thrown straight up, so its entire initial speed contributes to gaining height.

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Range R = (u^2 sin 2(theta))/g is maximum when sin 2(theta) = 1, i.e. theta = 45 degrees, giving R_max = u^2/g.

For a projectile launched with speed u at angle theta to the horizontal, the range is:

R = u^2 sin(2 theta) / g.

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