Question

A body is dropped from the top of a tower of height H. Average velocity of the body
is:
1)root (gh)
2)root(2gH)
3) root(2gH)/2
4)2root(gh)

​

Answer 1

• Option (3)
• √(2gH)/2

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$\orange{\bigstar}$  Correct Question  $\green{\bigstar}$

A body is dropped from the top of a tower of height H. Average velocity of the body  is :

1 . √(gH)

2 . √(2gH)

3 . √(2gH)/2

4 . 2√(gH)

$\orange{\bigstar}$  Given  $\green{\bigstar}$

A body is dropped from the top of a tower of height H

$\orange{\bigstar}$  To Find  $\green{\bigstar}$

Average velocity of the body

$\orange{\bigstar}$  Concept Used  $\green{\bigstar}$

→ 3rd Equation of motion

→ v² - u² = 2as

where ,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration due to gravity
• s denotes height

$\orange{\bigstar}$  Solution  $\green{\bigstar}$

Initial velocity , u = 0 m/s

Since , Dropped from height

Height of the tower , s = H m

Final velocity , v = v m/s

Acceleration due to gravity , a = g m/s²

Apply formula ,

$\to \bf v^2-u^2=2as\\\\\to \rm v^2-(0)^2=2(g)H\\\\\to \rm v^2=2gH\\\\\to \bf v=\sqrt{2gH}\ \;$

So , Final Velocity of the body is √(2gH)

Now , Average velocity is given by ,

$\bf \dfrac{u+v}{2}\\\\\to \rm \dfrac{0+\sqrt{2gH}}{2}\\\\\to \bf \dfrac{\sqrt{2gH}}{2}\ \; \pink{\bigstar}$

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Answer 2

Given :-

• A body is dropped from the top of a tower of height H

To Find :-

• Average velocity of the body

Formula :-

We know that,

→ 3rd Equation of motion

• v² - u² = 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Height

Apply formula,

• v² - u² = 2as

Substitute the values we get,

$\red{\implies \sf v^2-(0)^2=2(g)H}\\\\\green{\implies \sf v^2=2gH}\\\\\pink{\implies \sf v=\sqrt{2gH}}$

$\large{\gray{\therefore\underline{\sf Final \: Velocity \: of \: the \: body \: is \: \sqrt{2gH}}}}$

Now, Average velocity is given by,

$\blue{\sf \dfrac{u+v}{2}}\\\\\red{\implies \sf \dfrac{0+\sqrt{2gH}}{2}}\\\\\green{\implies\boxed{\sf\dfrac{\sqrt{2gH}}{2}}}$

Correct Option :-

$\green{3)\:\boxed{\sf\dfrac{\sqrt{2gH}}{2}}}$