Q. A projectile is thrown at angle *β* with vertical. It reaches a maximum height H. the time it takes to reach the highest point of its path is

(a) $ \displaystyle \sqrt \frac{H}{g} $

(b) $ \displaystyle \sqrt \frac{2 H}{g} $

(c) $ \displaystyle \sqrt \frac{H}{2 g} $

(d) $ \displaystyle \sqrt \frac{2 H}{g cos\beta} $

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Sol: $\large x = u sin\beta t$ …(i)

$\large y = u cos\beta t – \frac{1}{2}g t^2$ …(ii)

Time taken to reach highest Point is $t_H = \frac{t}{2} = \frac{u cos\beta}{g}$

From (ii)

$\large H = u cos\beta (\frac{u cos\beta}{g}) – \frac{1}{2}g (\frac{u cos\beta}{g})^2$

$\large H = \frac{u^2 cos^2 \beta}{2 g} $

$\large H = (\frac{u cos\beta}{g})^2 \times \frac{g}{2}$

$\large H = \frac{t_H^2 g}{2}$

$\large t_H = \sqrt{\frac{2 H}{g}}$