Q. A body is projected vertically upwards with a velocity ‘*u*’. It crosses a point in its journey at a height ‘*h*’ twice, just after 1 and 7 seconds. The value of u in ms^{-1 }is (*g* = 10*ms*^{-2})

(a) 50

(b) 40

(c) 30

(d) 20

Ans: (b)

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} $

1+7 = 2u/g

u = 4g = 40 m/s