A body is thrown vertically upward at t = 0. | vedclass

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Question

(A) The equation of motion for a body thrown vertically upward is given by $s = ut + \frac{1}{2}at^2$.Substituting $s = h$ and $a = -g$,we get $h = ut

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$t_1 + t_2 = t'_1 + t'_2$

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(A) The equation of motion for a body thrown vertically upward is given by $s = ut + \frac{1}{2}at^2$.Substituting $s = h$ and $a = -g$,we get $h = ut - \frac{1}{2}gt^2$.Rearranging this into a quadratic equation in $t$: $\frac{1}{2}gt^2 - ut + h = 0$,or $t^2 - \frac{2u}{g}t + \frac{2h}{g} = 0$.For a given height $h$,the roots of this quadratic equation are the two time instants $t_1$ and $t_2$ at which the body is at that height.According to the properties of quadratic equations,the sum of the roots is given by the coefficient of $t$ with a negative sign.Therefore,$t_1 + t_2 = \frac{2u}{g}$.Since this sum depends only on the initial velocity $u$ and acceleration due to gravity $g$,and not on the height $h$,the sum of the time instants for any height $h$ will be the same.Thus,$t_1 + t_2 = t'_1 + t'_2$.

$A$ body is thrown vertically upward at $t = 0$. It is at a height $80 \ m$ at instants $t_1$ and $t_2$. Also,it is at a height $60 \ m$ at instants $t'_1$ and $t'_2$. Then:

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