A body is falling freely from a point A at a | vedclass

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(B) Let the velocity of the particle at point $B$ be $v$. Since the body is falling freely, the acceleration is $g$. For the interval $BC$, the time t

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$0.5$

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(B) Let the velocity of the particle at point $B$ be $v$. Since the body is falling freely, the acceleration is $g$. For the interval $BC$, the time taken is $t_1 = 2 \ s$. Using the equation of motion $s = ut + \frac{1}{2}at^2$: $BC = v(2) + \frac{1}{2}g(2)^2 = 2v + 2g$  --- $(i)$ For the interval $BD$, the total time is $t_2 = 2 + 1 = 3 \ s$. $BD = BC + CD = 2BC$ (since $BC = CD$). $BD = v(3) + \frac{1}{2}g(3)^2 = 3v + 4.5g$ Since $BD = 2BC$, we have: $2(2v + 2g) = 3v + 4.5g$ $4v + 4g = 3v + 4.5g$ $v = 0.5g$ Now, for the interval $AB$, let the time taken be $t$. Since the body falls freely from $A$, its initial velocity $u = 0$. $v = u + at \Rightarrow 0.5g = 0 + gt$ $t = 0.5 \ s$.

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