A stone is dropped from the top of the tower of height 100m. the stone penetrates in the sand on the ground through distance of 2m. calculate the retardation of the stone.
Answers
Answer:
- Magnitude of Retardation of stone = 490 m/s²
Explanation:
Given:
- Stone is dropped from the top of the tower so Initial velocity = 0
- Height of tower = 100 m
- Stone penetrates in the sand on the ground through a distance = 2 m
- We will take acceleration due gravity during the motion = 9.8 m/s²
To find:
- Retardation of the stone =?
Formula required:
Third equation of motion
- 2 a s = v² - u²
[ Where v is representing final velocity, u is representing initial velocity, a is acceleration and s is distance covered ]
Solution:
Using third equation of motion to calculate final velocity of the stone when it will reach the ground (v=?)
[ Taking initial velocity zero; final velocity v; acceleration equal to acceleration due to gravity; and distance covered 100 m ]
→ 2 a s = v² - u²
→ 2 ( 9.8 ) ( 100 ) = v² - ( 0 )²
→ v² = 1960
→ v = √1960 m/s
Now, Using third equation of motion to calculate retardation of the stone when it penetrates in the sand.
[ Taking initial velocity equal to Final velocity of stone when it reaches the ground that is √1960 m/s; Taking final velocity equal to zero in this case because stones is coming to rest; Let, acceleration of stone be a ; and distance covered 2 m ]
→ 2 a s = v² - u²
→ 2 ( a ) ( 2 ) = ( 0 )² - ( √1960 )²
→ 4 a = -1960
→ a = -490 m/s²
Therefore,
- Acceleration of stone would be -490 m/s².
[ Negative sign represent that stone is retarding ]
Answer:
Answer is 490m/s.................