A man in a balloon, throws a stone downwards with a speed of 5 m/s with respect to balloon. The balloon is moving upwards with a constant acceleration of \[5\text{ }m{{s}^{-2}}\] . Then velocity of the stone relative to the man after 2 second is :
A 10 m/s
B 30 m/s
C 15 m/s
D 35 m/s

Hint: This question can be easily solved by relative motion. First of all we will find the velocity of the stone corresponding to the balloon after 2 seconds. Then we will find the relative velocity of stone with respect to the man by adding both the velocity because both are moving towards each other which result in increasing the relative velocity.
Formula used-
\[v=at+u\quad \]
V= final velocity
A = acceleration
U = initial velocity
T = time

Complete answer:
We construct three conventional equations of motion, commonly known as the laws of constant acceleration, in the situation of motion with uniform or constant acceleration (one with equal change in velocity in equal intervals of time). These equations regulate the motion of a particle by including the parameters displacement(s), velocity (initial and final), time(t), and acceleration(a). Only when a body's acceleration is constant and motion is on a straight line can these equations be used. The formula we'll use is v = u + at.
The acceleration of an item in free fall within a vacuum is known as gravitational acceleration in physics (and thus without experiencing drag). This is the gradual increase in speed induced only by gravitational pull. All bodies accelerate in vacuum at the same rate at a particular place on the Earth's surface, regardless of their mass or composition; gravimetry is the measurement and study of these speeds.
Given Initial speed of the balloon is $\mathbf{u}=\mathbf{5}~\mathbf{m}{{\text{s}}^{-1}}$
After 2 second
$\mathbf{v}=\mathbf{u}+\mathbf{gt}=\mathbf{5}+\mathbf{10}\times \mathbf{20}=\mathbf{25}~\mathbf{m}{{\mathbf{s}}^{-1}}$
after 2 second the speed of balloon will become $\text{v}=\dfrac{1}{2}\text{a}{{(2)}^{2}}=2\text{a}=2\times 5=10~\text{m}{{\text{s}}^{-1}}$
Since the Balloon is upward and the Stone is moving in downward directions (I e the opposite direction) so speed should be added quantitatively
Hence the relative speed $=\mathbf{2 5}+\mathbf{1 0}$ $=\mathbf{35}~\mathbf{m}{{\text{s}}^{-1}}$
$v=35m{{s}^{-1}}$

Hence option d is correct.

Note:
“In contrast to” is what “relative” means. When two or more bodies traveling at different speeds are examined, the notion of relative speed is employed. To make things easier, one body can be made stationary (i.e. Speed = 0), and the other body's speed with regard to the stationary body can be calculated as the total of the speeds if the bodies are travelling in opposing directions, or as the difference if they are going in the same direction. The relative Speed is the speed of a moving body in relation to a stationary body.