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Question

(C) Let the body be projected with initial velocity $u$. The equation of motion for height $h$ is $h = ut - \frac{1}{2}gt^2$, which is a quadratic equ

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 for height $h$ is $h = ut - \frac{1}{2}gt^2$, which is a quadratic equation in $t$: $\frac{1}{2}gt^2 - ut + h = 0$.Since $t_1$ and $t_2$ are the roots of this equation, the sum of roots is $t_1 + t_2 = \frac{u}{g/2} = \frac{2u}{g}$.Thus, $u = \frac{g(t_1+t_2)}{2}$.The maximum height $H$ reached by the body is given by $H = \frac{u^2}{2g}$.Substituting the value of $u$: $H = \frac{1}{2g} \left[ \frac{g(t_1+t_2)}{2} \right]^2 = \frac{1}{2g} \cdot \frac{g^2(t_1+t_2)^2}{4} = \frac{g(t_1+t_2)^2}{8}$.Alternatively, $H = 2g \left( \frac{t_1+t_2}{4} \right)^2 = 2g \cdot \frac{(t_1+t_2)^2}{16} = \frac{g(t_1+t_2)^2}{8}$.Therefore, the correct option is $C$.

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

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