Question

A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the position of the third, fourth, and fifth ball when the sixth ball is being dropped.
(A) $24.1m$ , $19.6m$ and $4.9m$ above the top.
(B) $44.1m$ , $19.6m$ and $4.9m$ below the top.
(C) $44.1m$ , $12.6m$ and $4.9m$ below the top.
(D) $41.4m$ , $29.6m$ and $4.9m$ below the top.

Answer 1

Hint Use Newton’s equations of motion to find the position of the third, fourth, and fifth ball. As the person is dropping the ball at regular intervals, take the initial velocity to be zero and take the acceleration to be the acceleration due to gravity.

Complete Step by step solution
Newton’s equation of motion to find the distance traveled by an object is given by
$s = ut + \dfrac{1}{2}a{t^2}$
Here, $u$ is the initial velocity, $t$ is the time taken to attain the final velocity, or it can also be defined as the time taken to travel the distance $s$ and $a$ is the acceleration due to gravity.
In this case, as the ball is being dropped and the ball falls under gravity, the acceleration will be the acceleration due to gravity. Substituting that into the equation gives us
$s = ut + \dfrac{1}{2}g{t^2}$
As the body is dropped, the initial velocity is always zero. Therefore,
$s = \dfrac{1}{2}g{t^2}$
By the time the sixth ball is dropped, six seconds have already passed. So the time passed when the sixth ball has been dropped for the third ball will be $3\sec $ , for the fourth ball will be $2\sec $ and for the fifth ball will be $1\sec $ .
Therefore, the position of the third ball will be computed with time $t = 3\sec $ .
${s_3} = \dfrac{1}{2} \times 9.8 \times {3^2}$
$ \Rightarrow {s_3} = 44.1m$
The position of the fourth ball will be computed with time $t = 2\sec $ .
${s_4} = \dfrac{1}{2} \times 9.8 \times {2^2}$
$ \Rightarrow {s_4} = 19.6m$
The position of the fifth ball will be computed with time $t = 1\sec $ .
${s_5} = \dfrac{1}{2} \times 9.8 \times {1^2}$
$ \Rightarrow {s_5} = 4.9m$
Therefore, the distance traveled by the third, the fourth, and the fifth ball by the time the sixth ball is dropped is $44.1m$ ,$19.6m$, and $4.9m$ respectively.

Hence, option (B) is the correct option.

Note
It should be noted that the distance that we have calculated is the distance from the top that the ball has traveled. It should not be confused with the distance above the ground. The solution can also be approached using the other two equations of motion but it will be longer compared to the usage of the equation we have used in this question.