A body falls from a height of 200 m. If the gravitation attraction ceases after 2 s, further time taken by it to reach the ground is ________. (g = 10 ms⁻²)

A body falls from a height of 200 m. If the gravitation attraction ceases after 2 s, further time taken by it to reach the ground is ________. (g = 10 ms⁻²)

a) 5 s               b) 9 s               c) 13 s             d) 17 s

From the given data we can conclude that:

1. The gravity acts on the object for the first 2 sec, then it ceases.

2. Then it reaches the ground with the velocity (velocity - at the end of 2 sec). 

3. We have to find the time it takes to reach the ground with velocity (that was after the end of 2 s)

Solution:

Given:

   u = 0

   v after 2 sec = ?

   a = 10 ms⁻²

To find:

The further time taken by the body to reach the ground. 

Sol:

We have to use the second equation of motion(as it is subjected to gravity) to find the distance it covered during the 2 s (during the time it was subjected to gravity).

Here, gravity = acceleration 

S = ut + 1/2(gt²)   at the end of 2 sec

S = 0(2) + 1/2(10 x 2 x 2)

S = 0 + 20

S = 20 m after 2 sec.

Now use the first equation of motion to find the velocity at the end of 2 s.

(acceleration = gravity)

Velocity = ?

v = u + gt    at the end of 2 sec

v = 0 + 10(2) 

v = 0 + 20 

v = 20 m/s at the end of 2 sec

Using the distance it traveled during 2 s, we can find the remaining distance.

Remaining distance = total distance - Distance traveled during 2 s

Remaining distance = 200 - 20

Remaining distance = 180 m

The remaining distance is 180 m.

As the body is not subjected to gravity after 2 s, we should find the time traveled by the body after 2 sec using the normal  "Distance, time, speed" formula

So,

t = d / v

t = (180 m) / (20 m/s)

t = 9 s

∴ A body falls from a height of 200 m. If the gravitation attraction ceases after 2 s, further time taken by it to reach the ground is     9 s      . (g = 10 ms⁻²)