A body of mass 500 g is thrown upwards with a velocity `20 ms^-1` and reaches back to the surface of a planet after `20 sec`. Then the weight of the body on that planet is (Assume g to be constant) xN. Find x.

Explanation

To solve the problem step by step, we will follow the principles of kinematics and the concept of weight.

### Step 1: Understand the problem
We have a body of mass 500 g (which is 0.5 kg) thrown upwards with an initial velocity of 20 m/s. It reaches back to the surface of the planet after 20 seconds. We need to find the weight of the body on that planet, which is given by the formula \( W = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity on that planet.

### Step 2: Determine the time of ascent and descent
Since the total time for the body to go up and come back down is 20 seconds, we can conclude that the time taken to ascend is equal to the time taken to descend. Therefore, the time of ascent \( t_a \) is:
\[ t_a = \frac{20 \text{ seconds}}{2} = 10 \text{ seconds} \]

### Step 3: Apply the kinematic equation
We can use the kinematic equation to find the acceleration due to gravity \( g' \) on the planet. The equation we will use is:
\[ v = u + at \]
Where:
- \( v \) is the final velocity (0 m/s at the highest point),
- \( u \) is the initial velocity (20 m/s),
- \( a \) is the acceleration (which will be \( -g' \) since it acts downward),
- \( t \) is the time of ascent (10 seconds).

Rearranging the equation gives us:
\[ 0 = 20 - g' \cdot 10 \]
Solving for \( g' \):
\[ g' \cdot 10 = 20 \implies g' = \frac{20}{10} = 2 \text{ m/s}^2 \]

### Step 4: Calculate the weight of the body
Now that we have the acceleration due to gravity \( g' \), we can calculate the weight \( W \) of the body using the formula:
\[ W = mg' \]
Where:
- \( m = 0.5 \text{ kg} \) (mass of the body),
- \( g' = 2 \text{ m/s}^2 \).

Substituting the values:
\[ W = 0.5 \text{ kg} \times 2 \text{ m/s}^2 = 1 \text{ N} \]

### Step 5: Conclusion
The weight of the body on that planet is \( 1 \text{ N} \). Therefore, the value of \( x \) is:
\[ x = 1 \]