Question

a body is allowed to fall from a tower 320 m high. when the ball reaches half the way acceleration due to gravity suddenly disappears. the velocity of the body after 2 seconds is
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Answer 1

Answer:

The Velocity of the body will be 0 m/s regardless of the distance it has covered. This is due to the sudden disappearing of gravitational force due to which the body will stop in the halfway and as there will be no force acting on it, final velocity will be 0 after 2 seconds.

Hope this helps.

Answer 2

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Question:-

A body is allowed to fall from a tower 320 m high. When the ball reaches half the way acceleration due to gravity suddenly disappears. The velocity of the body after two seconds is (g = 10 m/s).

To Find:-

Find the velocity of the body.

Solution:-

Given ,

A body is allowed to fall from a tower 320 m high.

If reaches half its distance 320/2 = 160 m.

First ,

We have to find the time travelled in half distance:-

$\dashrightarrow\sf \: s = ut + \dfrac { 1 } { 2 } { at }^{ 2 }$

$\dashrightarrow\sf \: 160 = \dfrac { 1 } { 2 } \times 10 \times { t }^{ 2 }$

$\dashrightarrow\sf \: 160 = 5 { t }^{ 2 }$

$\dashrightarrow\sf \: { t }^{ 2 } = \dfrac { 160 } { 5 }$

$\dashrightarrow\sf \: t = \sqrt { 32 } \: seconds$

How ,

We have to find the final velocity:-

$\dashrightarrow\sf \: v = u + at$

$\dashrightarrow\sf \: v = 0 + 10 \sqrt { 32 }$

$\dashrightarrow\sf \: v = 10 \sqrt { 32 } \: m/s$

Now ,

We have to find the velocity after 2 seconds:-

$\dashrightarrow\sf \: ( \sqrt { 32 } + 2 ) \: seconds = 10 \sqrt { 32 }$

$\dashrightarrow\sf \: 56 \: m/s$

Hence ,

Velocity after 2 seconds is 56 m/s.

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