Q. A stone projected vertically up from the ground reaches a height y in its path at t_{1} seconds and after further t_{2} seconds reaches the ground. The height y is equal to

(a) $ \displaystyle \frac{1}{2} g(t_1 + t_2)$

(b) $\displaystyle \frac{1}{2} g(t_1 + t_2)^2 $

(c) $ \displaystyle \frac{1}{2} g(t_1 . t_2)$

(d) $\displaystyle g(t_1 . t_2)$

Ans: (c)

Sol:

$ \displaystyle h= ut -\frac{1}{2}gt^2$

$ \displaystyle gt^2 -2ut + 2h = 0 $

$\displaystyle t_1 + t_2 = \frac{2u}{g} and \quad t_1 t_2 = \frac{2h}{g}$

Put h = y

$\displaystyle y = \frac{1}{2}g t_1 t_2 $