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

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(D) Let the particle be projected with an initial velocity $u$. The equation of motion for height $h$ is given by $h = ut - \frac{1}{2}gt^2$.Rearranging this,we get a quadratic equation in $t$: $\frac{1}{2}gt^2 - ut + h = 0$.Given that the particle reaches height $h$ at two times $t_1 = 5 \ s$ and $t_2 = 9 \ s$,these are the roots of the quadratic equation.From the properties of quadratic equations,the sum of the roots is given by $t_1 + t_2 = -\frac{b}{a} = \frac{u}{g/2} = \frac{2u}{g}$.Substituting the given values: $5 + 9 = \frac{2u}{10}$.$14 = \frac{2u}{10} \Rightarrow 2u = 140 \Rightarrow u = 70 \ m/s$.

$A$ particle is projected vertically upwards such that it attains a height of $h$ after $5 \ s$ and $9 \ s$ of its motion. The speed of projection is $...... \ m/s$ $(g = 10 \ m/s^2)$.

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