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