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

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Aicharya · Accepted Answer

The given situation is shown below

For Ist part of journey, $\begin{aligned} & u=0, a=-g=-10 \mathrm{~ms}^{-2} \\ & s=-40 \mathrm{~m}\end{aligned}$ Now using, $v^2-u^2=2 a s$ we get, $\quad v^2-0=2 \times(-10)(-40)$ $\Rightarrow \quad v^2=800$...(i) And using $v=u+a t$, we get; $\quad v=0+a t_1$, or $\quad v^2=a^2 t_1^2$ $\Rightarrow \quad t_1^2=\frac{v^2}{a^2}=\frac{800}{100}=8 \Rightarrow t_1=2 \sqrt{2}$ Now, for second part of journey, $u=0, a=-10 \mathrm{~m} / \mathrm{s}^2, s=-80 \mathrm{~m}$ $\Rightarrow v^2-u^2=2 a s \Rightarrow v^2=2 \times(-10)(-80)=1600$ And $\quad v=u+a t_2$ gives, $v=a t_2 \Rightarrow v^2=a^2 t_2^2$ Hence, $1600=100 \times t_2^2$ or $\quad t_2=4 \mathrm{~s}$ Time to complete last $50 \%$ part of the journey $=t_2-t_1=4-2 \sqrt{2} \simeq 1.17 \mathrm{~s}$

A ball is dropped from a tower of height $80 \mathrm{~m}$. The time it takes to cover the last $50 \%$ of…

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