Question

Two bodies one held 10 m vertically above the other,are released simultaneously. After falling freely for 3 seconds under gravity,their relative separation is freely falling body

Answer 1

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Answer 2

Answer:

After falling freely for 3 seconds under gravity, their relative separation is freely falling body is 10m.

Explanation:

Now, initial separation S=10m

Body 1:

Initial Speed(u₁)=0

Acceleration(g)=10m/s²

Time(t) =3 seconds

Now, we know that,

$\mathrm{s}=u t+\frac{1}{2} a \mathrm{t}^{2}$

So,

$\mathrm{s}_{1}=\frac{1}{2} \times 10 \times 9$

$\mathrm{s}_{1}=45 \mathrm{~m}$     (1)

Body 2:

Initial Speed(u₂)=0

Acceleration(g)=10m/s²

Time t=3 seconds

Now, we know that,

$\mathrm{s}=u \mathrm{t}+\frac{1}{2} a \mathrm{t}^{2}$

So,

$\mathrm{s}_{2}=\frac{1}{2} \times 10 \times 9$           (2)

Equations (1) and (2) result in the following;

S₁=S₂

As in the given time, the distance traveled by the body is the same, so the separation between them will not change .i.e.it will remain the same.

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