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} $
Click to See Answer :
Ans: (b)
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} $
$\large H = (\frac{u cos\beta}{g})^2 \times \frac{g}{2}$
$\large H = \frac{t_H^2 g}{2}$
$\large t_H = \sqrt{\frac{2 H}{g}}$