Question

Two bodies are released from the same height
at an interval of 1 s. How long after the first
body begins to fall will the two bodies be 10 m
apart ? (g = 10 m/s2)
[Ans. 1.5 s)​

Answer 1

The bodies travel in only one direction, that is the vertical direction. We have Acceleration due to gravity acting on them.

For the first body

Initial velocity = 0

Acceleration = g = 10 m/s²

Let distance be H in time t.

Distance traveled in time t is given by,

Using second equation of motion, S = ut + 1/2gt²

H = 0(t) + 1/2gt²

H = 5t²

For the second body,

Initial velocity = 0

Acceleration = g = 10 m/s²

Let distance traveled by S in time( t - 1) seconds.

Here We take t-1 as the second body is dropped after a second after releasing first one.

Using second equation of motion, S = ut + 1/2gt²

S= 0(t-1) + 1/2g(t-1)²

S= 5(t-1)²

According to the question, We need to find the time after the release of first ball the two bodies will be separated by 10m.

Since, first travels more distance than second due to time gap of release,

We can write as

H = S + 10

Substituting the values,

5t² = 5(t-1)² + 10

5t² = 5 ( t²+1-2t)+10

5t² = 5t² + 5 - 10 t + 10

- 10t + 15 = 0

-10t = - 15

t = 15/10 = 1.5 seconds.

Therefore, 1.5 seconds after the first

Therefore, 1.5 seconds after the firstbody begins to fall will the two bodies be 10 m apart

Answer 2

Explanation:

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