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# Two balls are dropped from different heights at different instants. Second ball is dropped $2\sec$ after the first ball. If both balls reach the ground simultaneously after $5\sec$ of dropping the first ball, the difference of initial heights of the two balls will be $(g = 9.8m/{s^2})$A. $58.8m$B. $78.4m$C. $98.0m$D. $117.6m$

Last updated date: 11th Aug 2024
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Hint:To solve this question, we have to know about acceleration due to gravity. The increasing speed which is acquired by an item due to gravitational power is called its acceleration due to gravity. The increasing speed because of gravity at the outside of Earth is spoken to by the letter g. It has a standard worth characterized as 9.80665 ${m/s^2}$.

According to the question, we are assuming, the first ball will be at height ${h_1}$ and it will take ${t_1}$ time to touch the ground which is $5\sec$. And in starting the second ball we can assume it will be at the height ${h_2}$ and it will take ${t_2}$ time to touch the ground, which is $5 - 2 = 3$ second.
Distance travelled by an ball falling for time $= h = \dfrac{1}{2}g{t^2}$
So, we can say, ${h_1} = \dfrac{1}{2}g{(5)^2} = 25g/2$
And again, ${h_2} = \dfrac{1}{2}g{(3)^2} = 9g/2$
${h_1} - {h_2} = 25g/2 - 9g/2 \Rightarrow{h_1} - {h_2}= 8 \times 9.8 \\ \therefore{h_1} - {h_2}= 78.4m \\$