Question

One body is dropped while a second body is thrown downwards with an initial velocity of 2m/s simultaneously. The separation between them is 18metres after a time : ?​

Answer 1

Solution:

Suppose:

After t second the separation will be 18 m.

In t second the first ball will travel by a distance equal to:

$\sf{d_{1} =\frac{1}{2}\:gt^{2}}$

$\sf{d_{1} =\frac{1}{2} \times 9.8\times t^{2}}$

Now:

The second ball will travel by distance:

$\sf{d_{2} = ut + \frac{1}{2}\:gt^{2}}$

$\sf{d_{2} = 2 \times t + \frac{1}{2}\times 9.8 \times t^{2}}$

So,

The separation is:

$\sf{d_{2} - d_{1} = \frac{1}{2} \times 9.8 \times t^{2} - (2t + \frac{1}{2} \times 9.8 \times t^{2})}$

$\implies$ 18 = 2 × t

$\implies$ t = 18/2

$\implies$ t = 9

Hence:

Separation will be equal to 18 m after 9 sec.

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Answered by: Niki Swar, Goa❤️

Answer 2

Answer:

The separation between two bodies will be 18m after 9 seconds.

Explanation:

Case-I: When one body is dropped. It means initially body was at rest. Therefore initial velocity, $u=0$. The body falls under the action of gravity ∴ $a=g$

From the second equation of motion:

$S=ut+\frac{1}{2}at^{2}$

Therefore, $S_{1}=(0)t+\frac{1}{2}gt^{2}$      

         $S_{1}=\frac{1}{2}gt^{2}$                                                           ...............(1)

Case-II: When a 2nd body is thrown with an initial velocity of 2m/s downwards.

Here, $u=2ms^{-1}$

Then, $S_{2}=(2)t+\frac{1}{2}gt^{2}$                                              ...............(2)

If the both bodies are thrown at the same time, the separation between them will be 18m.

Subtract eq.(1) from eq.(2) to find time;

$S_{2}-S_{1}=(2t+\frac{1}{2}gt^{2} )-(\frac{1}{2}gt^{2} )$

$18m=2ms^{-1}*t$

$t=9sec$

Therefore, after 9sec the distance between these two bodies will be 18m.