A ball is dropped from a tower of height 80 , | vedclass

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(B) Total height $H = 80 \,m$. The last $50 \%$ of the fall corresponds to the distance from $40 \,m$ to $80 \,m$ from the top.Let $t_1$ be the time t

Vedclass · Accepted Answer

$1.17 \,s$

Vedclass · Answer

(B) Total height $H = 80 \,m$. The last $50 \%$ of the fall corresponds to the distance from $40 \,m$ to $80 \,m$ from the top.Let $t_1$ be the time taken to fall the first $40 \,m$ (first $50 \%$ of the height).Using $s = ut + \frac{1}{2}at^2$ with $u = 0$ and $a = 10 \,ms^{-2}$:$40 = 0 + \frac{1}{2} 	imes 10 	imes t_1^2$$40 = 5 t_1^2 \Rightarrow t_1^2 = 8 \Rightarrow t_1 = \sqrt{8} = 2 \sqrt{2} \,s \approx 2.828 \,s$.Let $t_2$ be the total time taken to fall the entire height of $80 \,m$.$80 = 0 + \frac{1}{2} 	imes 10 	imes t_2^2$$80 = 5 t_2^2 \Rightarrow t_2^2 = 16 \Rightarrow t_2 = 4 \,s$.The time taken to cover the last $50 \%$ of the fall is $\Delta t = t_2 - t_1$.$\Delta t = 4 - 2 \sqrt{2} = 4 - 2.828 = 1.172 \,s \approx 1.17 \,s$.

$A$ ball is dropped from a tower of height $80 \,m$. The time it takes to cover the last $50 \%$ of its fall is (acceleration due to gravity $= 10 \,ms^{-2}$)

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