Introduction to Projectile Motion
Projectile motion occurs when an object is thrown into the air near the Earth's surface and moves along a curved path under the action of gravity only.
Let a stone be projected with velocity $u$ at an angle $\theta$ with the horizontal. The initial velocity components are:
- Horizontal component: $u_x = u \cos \theta$
- Vertical component: $u_y = u \sin \theta$
1. Time of Flight ($T$)
The total time for which the stone remains in the air. At the highest point, vertical displacement is zero over the full flight path. Using $s = ut + \frac{1}{2}at^2$ in the y-direction: $0 = (u \sin \theta)T - \frac{1}{2}gT^2$ $T = \frac{2u \sin \theta}{g}$
2. Maximum Height ($H$)
At maximum height, the vertical velocity component becomes zero ($v_y = 0$). Using $v_y^2 = u_y^2 + 2a_y s_y$: $0 = (u \sin \theta)^2 - 2gH$ $H = \frac{u^2 \sin^2 \theta}{2g}$
3. Horizontal Range ($R$)
The horizontal distance traveled is the horizontal velocity multiplied by the time of flight: $R = u_x \times T = (u \cos \theta) \times (\frac{2u \sin \theta}{g})$ $R = \frac{u^2 (2 \sin \theta \cos \theta)}{g}$ Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$: $R = \frac{u^2 \sin 2\theta}{g}$
Condition for Maximum Range
For $R$ to be maximum, $\sin 2\theta$ must be maximum. The maximum value of $\sin 2\theta = 1$ when $2\theta = 90^\circ$, meaning $\theta = 45^\circ$. Thus, the maximum range is $R_{max} = \frac{u^2}{g}$.