From a place where, $g = 9.8 m/s^2$, a stone is thrown upwards with a velocity of $4.9 m/s$. The time taken by the stone to return to the earth is:
A) 2s
B) 1s
C) 4s
D) 8s

Hint: When an object is thrown vertically upward the gravitational force tends to attract it towards the earth thus bit opposes its upward movement and the acceleration due to gravity is taken negatively.
When a project is thrown at an angle its initial velocity can be resolved into two components $\left( {{U_x}} \right)$ along the horizontal direction and $\left( {{U_y}} \right)$ along the vertical direction

Complete step by step solution:
Projectile motion is a type of motion in which an object moves in both the direction upward as well as forward symmetrical and in a parabolic path. The path in which the object moves is called its trajectory. Projectile motion occurs only when there is some force applied at the beginning of the trajectory. After this initial thrust, the only interference is from gravity
Refer to the following diagram for the motion of the object

Here ${V_o}$is the initial velocity of the object
${V_{ox}}$=${V_o}\operatorname{Cos} \theta $component of velocity along the horizontal direction
${V_{oy}} = {V_o}\operatorname{Sin} \theta $ component along the vertical direction.
Time of Flight:
The time of flight of projectile motion is defined as the time from when the object is projected to the time it reaches the surface.TT depends on the initial velocity magnitude and the angle of the projectile. The time of flight of the object is given by,
\[
  T = \left( {\dfrac{{2{U_Y}}}{g}} \right) \\
  i.eT = \left( {\dfrac{{2U \times \sin {\theta _{}}}}{g}} \right) \\
 \]
Here ${U_y}$is the component of initial velocity in the vertical direction.
Here it is said that the object is thrown upward that means $\theta = 90^\circ $
It is given that
$
  \left( U \right) = 4.9\dfrac{m}{{\sec }} \\
\Rightarrow g = 9.8\dfrac{m}{{\sec }} \\
 $
Now substituting the values get
\[
  \because T = \left( {\dfrac{{2U \times \sin \theta }}{g}} \right) \\
  \therefore T = \left( {\dfrac{{2 \times 4.9 \times \sin 90^\circ }}{{9.8}}} \right) \\
   \Rightarrow T = 1\sec \\
 \]
Thus, the total time of flight is \[T = 1\sec \].

Final answer is (B), the total time taken by the object o reach to the earth will be \[T = 1\sec \].

Note: Objects that are projected from and land on the same horizontal surface will have a vertically symmetrical path.
The time it takes for an object to be projected and the land is called the time of flight. It depends on the initial velocity of the projectile and the angle of projection.
When the projectile reaches a vertical velocity of zero, this is the maximum height of the projectile and then gravity exceeds all the forces and will accelerate the object downward.
The horizontal displacement of the projectile is called the range of the projectile and depends on the initial velocity of the object.