Projectile Motion

1. The horizontal motion has constant velocity (no horizontal acceleration).
2. The vertical motion has constant acceleration due to gravity $g = 9.8\ \text{m/s}^2$.
3. The motion is symmetrical: the time to rise is equal to the time to fall back.

Let a projectile be launched with an initial velocity $u$ at an angle $\theta$ with the horizontal.

Then:

Horizontal component: $u_x = u \cos \theta$

Vertical component: $u_y = u \sin \theta$

Step 1: Resolve Initial Velocity into Components

Let an object be projected with initial velocity $u$ at an angle $\theta$ from the horizontal.

Horizontal component: $u_x = u \cos \theta$

Vertical component: $u_y = u \sin \theta$

Step 2: Vertical Velocity After Time $t$

Using the first equation of motion vertically: $v_y = u_y + a_y t = u \sin \theta - g t$

Step 3: Horizontal Velocity After Time $t$

There is no horizontal acceleration, so: $v_x = u \cos \theta$

Step 4: Vertical Displacement After Time $t$

Using the second equation of motion: $s_y = u_y t + \frac{1}{2} a_y t^2$ $s_y = (u \sin \theta)t - \frac{1}{2} g t^2$

Step 5: Horizontal Displacement After Time $t$

Using the second equation of motion horizontally: $s_x = u_x t + \frac{1}{2} a_x t^2$

Since $a_x = 0$, this simplifies to: $s_x = (u \cos \theta)t$

Step 6: Maximum Height $H$

At maximum height, vertical velocity becomes zero: $v_y = 0$

Use the third equation of motion: $v_y^2 = u_y^2 + 2 a_y s_y$ $0 = (u \sin \theta)^2 - 2gH$

Solving for $H$: $H = \frac{u^2 \sin^2 \theta}{2g}$

Step 7: Time to Reach Maximum Height

$v_y = u \sin \theta - g t$

Set $v_y = 0$: $0 = u \sin \theta - g t$ $t = \frac{u \sin \theta}{g}$

Step 8: Total Time of Flight $T$

Since time to go up = time to come down: $T = 2 \times \frac{u \sin \theta}{g} = \frac{2u \sin \theta}{g}$

Step 9: Range $R$

Use horizontal displacement with total time of flight: $R = u \cos \theta \cdot T = u \cos \theta \cdot \frac{2u \sin \theta}{g}$ $R = \frac{u^2 \sin 2\theta}{g}$

Projectile motion involves the motion of an object projected into the air, moving under the influence of gravity only. The key parameters include:

(a) Trajectory

The trajectory is the curved path followed by the projectile. It is parabolic in nature. The trajectory equation is:

$y = x \tan\theta - \frac{g x^2}{2u^2 \cos^2\theta}$

Where:

• $y$: vertical displacement
• $x$: horizontal displacement
• $\theta$: angle of projection
• $u$: initial velocity
• $g$: acceleration due to gravity

(b) Time of Flight (T)

Total time the projectile is in the air:

$T = \frac{2u \sin\theta}{g}$

(c) Maximum Height (H)

Maximum vertical height reached by the projectile:

$H = \frac{u^2 \sin^2\theta}{2g}$

(d) Time to Reach Maximum Height (t)

Time taken to reach the highest point of the path:

$t = \frac{u \sin\theta}{g}$

(e) Horizontal Range (R)

Total horizontal distance traveled by the projectile:

$R = \frac{u^2 \sin 2\theta}{g}$

(f) Velocity at any point (v)

The magnitude of the velocity at any point along the path is given by:

$v = \sqrt{(u \cos\theta)^2 + (u \sin\theta - g t)^2}$

The direction (angle $\alpha$) of velocity at any point is:

$\alpha = \tan^{-1} \left(\frac{v_y}{v_x}\right)$

Where:

• $u$: initial velocity (m/s)
• $\theta$: angle of projection (degrees or radians)
• $g$: acceleration due to gravity (9.8 m/s²)
• $v_x = u \cos\theta$: horizontal velocity
• $v_y = u \sin\theta - g t$: vertical velocity
• $v$: magnitude of resultant velocity
• $\alpha$: angle of direction of velocity

Note: Maximum range is achieved when $\theta = 45^\circ$, and maximum height is achieved when $\theta = 90^\circ$.