Ball-1 is dropped from the top of a building | vedclass

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Question

(A) Let the balls meet at a distance $h$ from the top after time $t$. For ball-$1$ (downward motion): $h = u_1 t + \frac{1}{2} g t^2 = 0 \cdot t + \fr

Vedclass · Accepted Answer

$\frac{45}{4} \,m$

Vedclass · Answer

(A) Let the balls meet at a distance $h$ from the top after time $t$. For ball-$1$ (downward motion): $h = u_1 t + \frac{1}{2} g t^2 = 0 \cdot t + \frac{1}{2} (10) t^2 = 5 t^2$ ... $(i)$ For ball-$2$ (upward motion): The distance covered by ball-$2$ is $(21 - h)$. $(21 - h) = u_2 t - \frac{1}{2} g t^2 = 14 t - 5 t^2$ ... (ii) Adding equations $(i)$ and (ii): $h + (21 - h) = 5 t^2 + 14 t - 5 t^2$ $21 = 14 t$ $t = \frac{21}{14} = 1.5 \,s$ Substituting $t = 1.5 \,s$ into equation $(i)$: $h = 5 	imes (1.5)^2 = 5 	imes 2.25 = 11.25 \,m = \frac{45}{4} \,m$ Thus, ball-$1$ will have dropped $\frac{45}{4} \,m$ when it passes ball-$2$.

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