Question

A man is standing on top of a building \[100m\] high. He throws two balls vertically, one at \[t = 0\] and other after a time interval (less than \[2s\]). The later ball is thrown at a velocity of half the first. The vertical gap between the first and second ball is \[15m\] at \[t = 2s\]. The gap is found to remain constant. The velocities with which the balls were thrown are (Take \[g = 10m{s^{ - 2}}\]).
A. \[20m{s^{ - 1}},10m{s^{ - 1}}\]
B. \[10m{s^{ - 1}},5m{s^{ - 1}}\]
C. \[16m{s^{ - 1}},8m{s^{ - 1}}\]
D. \[30m{s^{ - 1}},15m{s^{ - 1}}\]

Answer 1

Hint: When an object is thrown vertically, it will eventually fall back to the ground under the earth's gravity. In fact all objects near the earth's surface fall with a constant acceleration of about \[9.8m{s^{ - 2}}\] (we usually take \[10m{s^{ - 2}}\] for the sake of simplicity in calculation). This is called the acceleration due to gravity and is usually denoted by the symbol.
The vertical motion under gravity can be described by the equations of motion that we have learned. The effect of air resistance is neglected in these examples.
In question, two balls are thrown in upward direction. Second ball is thrown after \[2s\].The gap between two balls will be considered the same in this question.

Complete step by step answer:
The only way for the gap to remain constant is if they have the same velocity for all times after that $2$ $seconds$. In fact, this must be true from the moment the second ball was thrown; otherwise their velocities would never be equal.

So if their velocities are equal at all times, that means when the man threw the second ball, he had to have given it an initial velocity equal to the first ball's velocity at that time. Also, since the distance is always \[15\]\[m\]the threw the second ball when the first was 15m away. And I almost forgot the second is thrown half as fast as the first.
Let, the initial velocity of the first ball=\[u\]\[m{s^{ - 1}}\].
The only possibility for the gap to remain constant is that: Both balls have the same velocity after the second ball is thrown.
When second ball is thrown, then Speed of first ball=\[\dfrac{u}{2}\]
For the first ball: Distance covered, \[S = 15m\], the initial velocity =\[u\]\[m{s^{ - 1}}\]. final velocity = \[\dfrac{u}{2}\]\[m{s^{ - 1}}\].
By Newton’s Third equation of motion, \[{v^2} = {u^2} - 2gS\]
Putting values, we get
\[ \Rightarrow {\left( {\dfrac{u}{2}} \right)^2} = {u^2} - 2 \times 10 \times 15\]
\[ \Rightarrow \dfrac{{3{u^2}}}{4} = 300\]
\[ \Rightarrow {u^2} = 400\]
\[ \Rightarrow u = 20\]\[m{s^{ - 1}}\]
When second ball is thrown, then Speed of first ball=\[\dfrac{u}{2}\]
\[u = \dfrac{{20}}{2}\]\[m{s^{ - 1}}\]
\[u = 10\]\[m{s^{ - 1}}\]
Therefore, Velocity of the first ball is \[20\]\[m{s^{ - 1}}\] in an upward direction. Velocity of the second ball is \[10\]\[m{s^{ - 1}}\] in upward direction.

So, the correct answer is “Option A”.

Note:
When two balls thrown in upward direction, students make mistakes of taking values of velocities, It may be noted that the first ball is thrown at \[t = 0\]. So we take reference of velocity of the first ball. When the second ball is thrown, that time velocities of two balls will be the same. At the time of the second ball thrown, the velocity of the first ball is taken as \[v\].