- Horizontal range.
- Maximum height attained.

$\therefore\ \text{R}=\text{u}\cos\theta_0\times$ Total time of flight (T)
$=\text{u}\cos\theta_0\times\frac{2\text{u}\sin\theta_0}{\text{g}}$
$=\frac{\text{u}^2}{\text{g}}2\sin\theta_0\cos\theta_0$
$=\frac{\text{u}^2}{\text{g}}\sin2\theta_0$
$\text{R}=\frac{\text{u}^2\sin2\theta_0}{\text{g}}$
Here, AH is referred as the maximum height attained against gravity. At highest point vertical component of velocity become zero.
Using, v2 = u2 + 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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$\Big(\text{Hint: acceleration}\frac{\text{V}^2}{\text{R}}=\frac{4\pi^2\text{R}}{\text{T}^2}\Big)$