This question was previously asked in

UP TGT Mathematics 2021 Official Paper

Option 1 : 9.8 m/s

__Concept:__

- Acceleration due to gravity (g): The acceleration applied due to the gravity of earth on any object moving upward or downward under influence of gravity is called acceleration due to gravity.
- Freefall motion under gravity: If an object is just dropped from a height, then it will fall down with acceleration due to gravity g.
- The upward motion under gravity: If the object is thrown with some velocity u in the upward direction, it is retarded by gravity. It reaches its maximum height where its velocity becomes zero. After that, it starts downward motion just as free fall.

- The time taken by the object to reach its maximum height is same as the time taken by the object to descent back. Equal distances are traveled in equal interval of time while ascending and descending.
- Equations of Motion under constant acceleration: The following three equations are considered under motion in straight line when there is a uniform acceleration.

v = u + at -- (1)

\(S = ut + \frac{1}{2}at^2\) -- (2)

v2 = u2 + 2as -- (3)

Since distance traveled by an object in the last t second of motion is equal to the distance traveled by the object first t second during its downward motion.

**Calculation:**

Given,

The stone is thrown up with velocity say u

time is taken by the ball to go up and come back t total = 6 second

Now, time of ascent = time of descent = half of the total time taken

So, time taken by the ball to reach its maximum height h is t u = 6 s / 2 = 3 s

At maximum height, the speed becomes zero, so final speed is v = 0

Now, since, here motion is against the acceleration due to gravity g, we take acceleration as

a = - g = - 9.8 m s^{ -2}

Now, applying this in first equation

v = u + at

⇒ 0 = u + (-9.8 m s ^{-2}) (3 s)

⇒ u = 29.4 m /s

So, the ball was thrown us with 29.4 m /s

Now, we have initial velocity u = 29.4 m /s

Acceleration a = - 9.8 m s ^{- 2}

We have to find speed after 2 seconds, t' = 2 s

Let speed after 2 seconds be v'

Now, applying the first equation again

v' = 29.4 m/s + (-9.8m) × (2 second)

⇒ v' = 29.4 m/s - 19.8 m/s = 9.8 m/s

**So, the speed after 2 second will be 9.8 m /s. **