Question

A stone is released from the top of a building 78.4m high. Its final velocity just before touching the ground is

Answer 1

Explanation:

GIVEN :-

Distance , s = 78.4m

Initial velocity , u = 0

final velocity , v = ?

According to Newton's third law of motion

===> v² = u² + 2gs

===> v² = ( 0 ) ² + 2 × 9.8 × 78.4

===> v² = 0 + 1536.64

===> v = √1536.64

===> v = 39.2 m/s

Here , g = acceleration due to gravity which is equal to 9.8 m/s .

Hence , final velocity , v is 39.2 m/s .

Answer 2

Answer :

›»› The final velocity of stone = 39.2 m/s

Given :

• Initial velocity of stone (u) = 0 m/s
• Stone dropped from height (h) = 78.4 m

To Find :

• Final velocity of stone (v) = ?

Required Solution :

→ Acceleration due to gravity (g) = 9.8 m/s (because stone is released from the top of a building)

From third equation of motion

⇒ v² = u² + 2gh

⇒ v² = 0² + 2 × 9.8 × 78.4

⇒ v² = 0 + 2 × 9.8 × 78.4

⇒ v² = 0 + 19.6 × 78.4

⇒ v² = 0 + 1536.64

⇒ v² = 1536.64

⇒ v = √1536.64

⇒ v = 39.2 m/s

║Hence, the Final velocity of stone is 39.2 m/s.║

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Additional Information :

★ First equation of motion :

⪼ v = u + gt

Where,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• g is the Acceleration due to gravity in m/s².
• t is the Time taken in second.

★ Second equation of motion :

⪼ h = ut + ½ gt²

Where,

• h is the height in m.
• u is the Initial velocity in m/s.
• t is the Time taken in second.
• g is the Acceleration due to gravity in m/s².

★ Third equation of motion :

⪼ v² = u² + 2gh

Where,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• g is the Acceleration due to gravity in m/s².
• h is the height in m.