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(A) The initial velocity $u = 0 \, m/s$ and the acceleration due to gravity is $g$. Using the equation of motion $s = ut + (1/2)gt^2$,for the total he

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$25 \, m$ from the top

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(A) The initial velocity $u = 0 \, m/s$ and the acceleration due to gravity is $g$. Using the equation of motion $s = ut + (1/2)gt^2$,for the total height $H = 100 \, m$ and time $T$: $100 = 0 \cdot T + (1/2)gT^2 \implies 100 = (1/2)gT^2 \implies T^2 = 200/g$. Now,let $h$ be the distance covered in time $t = T/2$: $h = (1/2)g(T/2)^2 = (1/2)g(T^2/4) = (1/8)gT^2$. Substituting $gT^2 = 200$ into the equation: $h = (1/8) \cdot 200 = 25 \, m$. Since $h$ is the distance covered from the top,the bottle will be $25 \, m$ from the top.

$A$ soft drink bottle is dropped by a waiter from the top of a hotel whose height is $100 \, m$. It takes $T$ seconds to reach the ground. Where will it be at $T/2$ seconds?

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