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(C) Let the body be projected with initial velocity $u$. The equation of motion is $h = ut - \frac{1}{2}gt^2$. Rearranging,we get $\frac{1}{2}gt^2 - u

Vedclass · Accepted Answer

$2g\left(\frac{t_1+t_2}{4}\right)^2$

Vedclass · Answer

(C) Let the body be projected with initial velocity $u$. The equation of motion is $h = ut - \frac{1}{2}gt^2$. Rearranging,we get $\frac{1}{2}gt^2 - ut + h = 0$. This is a quadratic equation in $t$ with roots $t_1$ and $t_2$. The sum of roots is $t_1 + t_2 = \frac{u}{g/2} = \frac{2u}{g}$,so $u = \frac{g(t_1+t_2)}{2}$. The maximum height $H$ reached by the body is given by $H = \frac{u^2}{2g}$. Substituting the value of $u$,we get $H = \frac{1}{2g} \left[ \frac{g(t_1+t_2)}{2} \right]^2 = \frac{1}{2g} \cdot \frac{g^2(t_1+t_2)^2}{4} = \frac{g(t_1+t_2)^2}{8}$. Alternatively,$H = 2g \left( \frac{t_1+t_2}{4} \right)^2 = 2g \cdot \frac{(t_1+t_2)^2}{16} = \frac{g(t_1+t_2)^2}{8}$. Thus,the correct option is $C$.

$A$ body projected vertically upwards crosses a point twice in its journey at a height $h$ just after $t_1$ and $t_2$ seconds. Maximum height reached by the body is

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