A ball projected from the ground vertically u | vedclass

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(B) Let the initial speed of projection be $u$. The equation of motion for the height $h$ at time $t$ is given by $h = ut - \frac{1}{2}gt^2$.Rearrangi

Vedclass · Accepted Answer

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

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(B) Let the initial speed of projection be $u$. The equation of motion for the height $h$ at time $t$ is given by $h = ut - \frac{1}{2}gt^2$.Rearranging this,we get a quadratic equation in $t$: $\frac{1}{2}gt^2 - ut + h = 0$.Since the ball is at the same height $h$ at times $t_1$ and $t_2$,these are the two roots of the quadratic equation.From the properties of quadratic equations,the sum of the roots is given by $t_1 + t_2 = -\frac{b}{a} = \frac{u}{g/2} = \frac{2u}{g}$.Solving for $u$,we get $u = \frac{g(t_1 + t_2)}{2}$.

$A$ ball projected from the ground vertically upward is at the same height at times $t_1$ and $t_2$. The speed of projection of the ball is [Neglect the effect of air resistance].

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