Question

A stone is released from the top of a tower of
height 19.6 m. Its final velocity just before
touching the ground is
(1) 9.8 m/s
(2) 19.6 m/s
(3) 4.9 m/s
(4) 10 m/s​

Answer 1

Answer:

Explanation:

Given,

Initial velocity of the stone, u = 0 m/s

Distance of the tower, s = 19.6 m

Acceleration due to gravity, g = 9.8 m/s²

To Find,

Final velocity of the stone, v = ?

Formula to be used,

3rd equation of motion,

v² - u² = 2as

Solution,

Putting all the values, we get

v² - u² = 2gs

⇒ v² - (0)² = 2 × 9.8 × 19.6

⇒ v² = 384.16

⇒ v = 19.6 m/s.

Hence, the final velocity just before  touching the ground is 19.6 m/s.

Answer 2

Answer:

Given :-

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

To Find :-

• What is the final velocity (v) just before touching the ground.

Formula Used :-

By using third equation of motion we know that,

✪ v² = u² + 2gs ✪

where,

• v = Final velocity
• u = Initial velocity
• g = Acceleration due to gravity
• s = Height

Solution :-

Given :

• Initial velocity = 0 m/s
• Height = 19.6 m
• Acceleration due to gravity = 9.8 m/s

According to the question by using the formula we get,

⇒ v² = u² + 2gs

⇒ v² = (0)² + 2(9.8)(19.6)

⇒ v² = 0 + 2(192.08)

⇒ v² = 0 + 384.16

⇒ v² = 384.16

⇒ v = √384.16

➠ v = 19.6 m/s

∴ The final velocity just before touching the ground is 19.6 m/s . Hence, the correct options is option no 2) 19.6 m/s.

✯ Extra Formula ✯

➤ First equation of motion :

● v = u + at

➤ Second equation of motion :

● s = ut + 1/2 at²

➤ Third equation of motion :

● v² = u² + 2as

where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time
• s = Distance covered