A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answers
Answered by
67
Answer:-
Given:-
Initial velocity of the stone = 0 m/s
Final velocity of the stone
Height of the stone = 19.6 m
Acceleration due to gravity = 9.8 ms−2
Solution:-
To find this we use, the equation of motion under gravity v²−u² = 2gs
∴ v²− 0²= 2 × 9.8 × 19.6
⇒ v² = 2 × 9.8 × 19.6 = (19.6)2
⇒ v² = √384.16
⇒ v = 19.6 ms−1
Hence, the velocity of the stone just before touching the ground is 19.6 ms−1 .
Answered by
59
Given:
Initial velocity of stone (u) = 0 m/S
Height of the tower (s) = 19.6 m
Acceleration due to gravity = 9.8 m/s²
To find:
We have to find the final velocity of stone just before touching the ground.
Solution:
To find the velocity, we will use third equation of motion:
v² - u² = 2gs
⇒ v² - 0² = 2 * 9.8 * 19.6
⇒ v² = 384.16
⇒ v = √384.16
⇒ v = 19.6 m
Hence, the final velocity of stone just before touching the ground- 19.6 m/s.
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