Question
A body is projected with velocity $u$ at an angle $\theta_0$ upward from horizontal. Deduce the expression for :
- Horizontal range.
- Maximum height attained.
✓
Answer
Consider a body is projected with velocity $u$ at an angle $\theta_0.$ The horizontal and vertical components initially with velocity are $\text{u}\cos\theta_0$ and $\text{u}\sin\theta_0$ respectively.
$=\text{u}\cos\theta_0\times\frac{2\text{u}\sin\theta_0}{\text{g}}$
$=\frac{\text{u}^2}{\text{g}}2\sin\theta_0\cos\theta_0$
$=\frac{\text{u}^2}{\text{g}}\sin2\theta_0$
$\text{R}=\frac{\text{u}^2\sin2\theta_0}{\text{g}}$
Using, $v^2 = u^2 + 2as$, we have
$0=\text{u}^2\sin^2\theta_0-2\text{gh}_\text{max}$
$\Rightarrow\ \text{h}_\text{max}=\frac{\text{u}^2\sin^2\theta_0}{2\text{g}}$
- Horizontal range: It is the horizontal distance covered by the body between its point of projection and the point of hitting the ground, when both points are on same horizontal plane. It is denoted by $R$.
$=\text{u}\cos\theta_0\times\frac{2\text{u}\sin\theta_0}{\text{g}}$
$=\frac{\text{u}^2}{\text{g}}2\sin\theta_0\cos\theta_0$
$=\frac{\text{u}^2}{\text{g}}\sin2\theta_0$
$\text{R}=\frac{\text{u}^2\sin2\theta_0}{\text{g}}$
- Maximum height : Maximum height ($h_{max}$) is the maximum vertical height attained by the body above the point of projection during its flight.
Using, $v^2 = u^2 + 2as$, we have
$0=\text{u}^2\sin^2\theta_0-2\text{gh}_\text{max}$
$\Rightarrow\ \text{h}_\text{max}=\frac{\text{u}^2\sin^2\theta_0}{2\text{g}}$
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