Question

10) A stone is released from the top of the tower
of height. 19.6m calculate the final velocity
just before touching the ground​

Answer 1

Given :

• A stone is released from the top of the tower of height 19.6m.

To Find :

• Final Velocity of the stone = ?

Solution :

• Height of the tower = 19.6 m
• Acceleration due to gravity = 9.8 m/s²
• Initial Velocity = 0 m/s

By using third equation of motion we get :

• v² - u² = 2as

Here,

• v is the Final Velocity of stone.
• u is the Initial Velocity of stone,
• s is the height of the tower,
• a is the acceleration due to gravity.

Now, plug in the given values in above equation of motion :

→ v² - 0² = 2(9.8)(19.6)

→ v² = 19.6 × 19.6

→ v² = 384.16

→ v = √384.16

→ v = 19.6 m/s

⛬ The Final Velocity of the stone would be 19.6 m/s.

Answer 2

$\large \underline{\underline{\frak{Given \: }}}$

• Height of the tower = 19.6 cm.
• u = 0m/s.
• G = 9.8m/s²

$\large \underline{\underline{\frak{To find \: }}}$

• Final velocity.

$\large \underline{\underline{\frak{solution \: }}}$

by using formula :-

$\huge \fbox \blue{v²=u²+2gh}$

Substituting the values in the formula,

$\implies$ v²=(0)²+2(9.8)(19.6)

$\implies$ v²= 384.16

$\implies$ v = √384.16 = 19.6m/s.

hence, the final velocity of the stone would be 19.6m/s.

$\large \underline{\underline{\frak{So, Done! \: }}}$