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  4. A projectile is thrown at angle β with vertical. It reaches a maximum height H . The time taken to reach the highest point of its path is

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Q. A projectile is thrown at angle $\beta$ with vertical. It reaches a maximum height $H .$ The time taken to reach the highest point of its path is

Motion in a Plane

Solution:

$t=\frac{u \sin (90-\beta)}{g}=\frac{u \cos \beta}{g}$
$H=\frac{u^{2} \sin ^{2}(90-\beta)}{2 g}=\frac{(u \cos \beta)^{2}}{2 g}$
$\Rightarrow u \cos \beta=\sqrt{2 g H}$
$\Rightarrow t=\frac{\sqrt{2 g H}}{g}=\sqrt{\frac{2H}{g}}$