(D) Let the body be projected with initial velocity $u$. The equation of motion for a height $h$ is given by $h = ut - \frac{1}{2}gt^2$. Rearranging t

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(D) Let the body be projected with initial velocity $u$. The equation of motion for a height $h$ is given by $h = ut - \frac{1}{2}gt^2$. Rearranging this,we get a quadratic equation in $t$: $\frac{1}{2}gt^2 - ut + h = 0$. The roots of this equation are $t_1$ and $t_2$,which represent the times at which the body reaches height $h$ during ascent and descent. From the properties of quadratic equations,the sum of the roots is $t_1 + t_2 = \frac{-(-u)}{\frac{1}{2}g} = \frac{2u}{g}$. Solving for $u$,we get $u = \frac{g(t_1 + t_2)}{2}$.

Class 11 Physics Motion in Straight Line Motion Under Gravity

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$A$ body is projected vertically upwards. The times corresponding to a specific height while ascending and while descending are $t_{1}$ and $t_{2}$ respectively. Then the velocity of projection is (where $g$ is acceleration due to gravity).

KCET 2008 All KCET

A
$g \sqrt{t_{1} t_{2}}$
B
$\frac{g t_{1} t_{2}}{t_{1}+t_{2}}$
C
$\frac{g \sqrt{t_{1} t_{2}}}{2}$
D
$\frac{g(t_{1}+t_{2})}{2}$

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$A$ body is projected vertically upwards. The times corresponding to a specific height while ascending and while descending are $t_{1}$ and $t_{2}$ respectively. Then the velocity of projection is (where $g$ is acceleration due to gravity).

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