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(C) Let $h = 5 \, m$ be the height and $\Delta t = 10 \, s$ be the time interval between the two passes.Using the equation of motion $h = ut - \frac{1

Vedclass · Accepted Answer

$\sqrt{104}$

Vedclass · Answer

(C) Let $h = 5 \, m$ be the height and $\Delta t = 10 \, s$ be the time interval between the two passes.Using the equation of motion $h = ut - \frac{1}{2}gt^2$,we can write $gt^2 - 2ut + 2h = 0$.Let $t_1$ and $t_2$ be the two times at which the object is at height $h$. These are the roots of the quadratic equation $t^2 - (2u/g)t + (2h/g) = 0$.The difference between the roots is $|t_2 - t_1| = \sqrt{(2u/g)^2 - 4(2h/g)} = 10$.Let $T = u/g$ be the time taken to reach the maximum height. Then $2T = 2u/g$.So,$\sqrt{(2T)^2 - 8h/g} = 10$.Squaring both sides: $4T^2 - 8h/g = 100$.Given $h = 5 \, m$ and $g = 10 \, m/s^2$,we have $4T^2 - 8(5)/10 = 100$.$4T^2 - 4 = 100 \Rightarrow 4T^2 = 104 \Rightarrow T^2 = 26$.$T = \sqrt{26} \, s$.The total time of flight is $2T = 2\sqrt{26} = \sqrt{4 	imes 26} = \sqrt{104} \, s$.

An object thrown vertically up from the ground passes a height of $5 \, m$ twice in an interval of $10 \, s$. The total time of flight is ........ $s$.

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