A body thrown vertically up with initial velo | vedclass

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(B) Let the initial velocity be $u = 52 \,m/s$ and the acceleration due to gravity be $g = 10 \,m/s^2$.Using the equation of motion $h = ut - \frac{1}

Vedclass · Accepted Answer

$10.2$

Vedclass · Answer

(B) Let the initial velocity be $u = 52 \,m/s$ and the acceleration due to gravity be $g = 10 \,m/s^2$.Using the equation of motion $h = ut - \frac{1}{2}gt^2$,we get $gt^2 - 2ut + 2h = 0$.This is a quadratic equation in $t$,having two roots $t_1$ and $t_2$,which represent the times at which the body is at height $h$.The sum of the roots is $t_1 + t_2 = \frac{2u}{g} = \frac{2 	imes 52}{10} = 10.4 \,s$.The difference of the roots is given as $t_2 - t_1 = 10 \,s$.Adding the two equations: $2t_2 = 20.4 \,s \Rightarrow t_2 = 10.2 \,s$.Subtracting the two equations: $2t_1 = 0.4 \,s \Rightarrow t_1 = 0.2 \,s$.The product of the roots is $t_1 t_2 = \frac{2h}{g}$.Substituting the values: $0.2 	imes 10.2 = \frac{2h}{10}$.$2.04 = \frac{h}{5} \Rightarrow h = 2.04 	imes 5 = 10.2 \,m$.

$A$ body thrown vertically up with initial velocity $52 \,m/s$ from the ground passes twice a point at $h$ height above at an interval of $10 \,s$. The height $h$ is .........$m$ $(g=10 \,m/s^2)$.

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