Question

Q: A body when released from a certain height reaches ground level in 4 seconds. What time does it take to cover half distance? Options: A) √2 seconds B) 2 seconds C) 2√2 seconds D) 2+√2 seconds

Answer 1

Answer:

2 seconds

Explanation:

to find height

Here, u = 0

t = 4 sec

a = 9.8m/s²

s = ?? ( we need to find)

By using,

S = ut + 1/2at²

Putting the values we get,

S = 0 + 1/2 * 9.8*(4)²

Distance (s) = 78.4 m

Answer 2

The time taken by the body to cover half the distance is 2√2 seconds (option-c).

Given: the time taken by the body to reach the ground, t₁ = 4s            

To Find: time taken by the body when it covers half distance, t₂.

Solution:

To calculate t₂ the formula used is:

• s = ut + 1/2 ( at²)

• here, s is the distance covered by the body

        u is the initial velocity of the body i.e. u= 0 m/s

        t is the time taken by the body to reach the ground

        a is the acceleration due to gravity, g = 9.8m/s²

First calculate s, by using the above formula:

s = u x t₁ + 1/2x a x t₁²

s = 0 x 4 + 1/2 x 9.8 x (4)²

  = 0 + 1/2 x 9.8 x 4x4

  = 1/2 x 9.8 x 16

  = 9.8 x 8

  = 78.4

s = 78.4 m

Now, when the distance is half then the formula used for t₂:

S = ut + 1/2 x a( t₂)²

here, S = s /2

s/2 = ut₂ +1/2 x a( t₂)²

78.4 /2 = 0xt₂ + 1/2 x 9.8 x ( t₂)²

39.2    = 0 + 1/2 x 9.8 x ( t₂)²

39.2    = 1/2 x 9.8 x ( t₂)²

39.2 x 2 =  9.8 x ( t₂)²

78.4  =  9.8 x ( t₂)²

78.4 / 9.8 = ( t₂)²

            8 = ( t₂)²

        ( t₂)²= 8

          t₂   = $\sqrt{8}$

                = √2x2x2

        t₂     = 2√2 s

Hence, the time taken by the body to cover half the distance is 2√2 seconds.