A particle is projected upwards. The times co | vedclass

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(D) Let the initial velocity of projection be $u$. The equation of motion for a particle at height $h$ is given by $h = ut - \frac{1}{2}gt^2$.Rearrang

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

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(D) Let the initial velocity of projection be $u$. The equation of motion for a particle at 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 particle is at height $h$ during ascent and descent.From the properties of quadratic equations,the sum of the roots is given by $t_1 + t_2 = -\frac{b}{a} = \frac{u}{1/2g} = \frac{2u}{g}$.Solving for the initial velocity $u$,we get $u = \frac{g(t_1 + t_2)}{2}$.

$A$ particle is projected upwards. The times corresponding to height $h$ while ascending and while descending are $t_1$ and $t_2$ respectively. The velocity of projection will be

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