Question

22. 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)

1 ) 80 m/s
2) 100 ms
3) 60 ms
4) 56 ms​

Answer 1

Given:-

• Initial velocity ,u = 0m/s
• Height of tower ,h = 320m (maximum height)
• Time taken ,t = 2s
• Acceleration due to gravity ,g = 10m/s²

To Find:-

• Velocity of body after 2 rsecond ,v

Solution:-

In 1st case AB

Firstly we calculate the time taken when body reach to B (refer to attachment )

• T = √2h/g

where,

• h denote height (when it reaches half )
• g denote acceleration due to gravity

Substitute the value we get

$:\implies$ T = √2×160/10

$:\implies$T = √320/10

$:\implies$T = √32 seconds

Therefore, final velocity = gt

$:\implies$ Final velocity = 10×√32

$:\implies$ Final Velocity = 10√32 m/s.

____________________

In 2nd case BC

In this case acceleration is not there .

Therefore, velocity will be Constant .

Velocity after (√32 + 2 )s = 10√32

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ = 56m/s

• Hence, the velocity after 2 second is 56m/s.

So , the required answer is (4) 56m/s.

___________________

Answer 2

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.