Question

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. \[117.6m\]

Answer 1

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 ${m/s^2}$.

Complete step by step answer:
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, \[{h_2} = \dfrac{1}{2}g{(3)^2} = 9g/2\]
So, the difference those heights is,
 \[{h_1} - {h_2} = 25g/2 - 9g/2
\Rightarrow{h_1} - {h_2}= 8 \times 9.8 \\
\therefore{h_1} - {h_2}= 78.4m \\ \]
So, the right option would be B.

Note:We can get confused between normal acceleration and acceleration due to gravitation. We have to know the definition of acceleration to solve other questions like this one. We know that acceleration is the rate of change of velocity. And velocity is the rate of change of speed. We have to keep that in our mind.