An Astronaut drops an object from a height of 2m and measures that it falls to the "ground" in 1.08 seconds. What is the value of "g" that can be calculated from this data?
Answers
Given:-
- Initial Velocity = 0m/s. (as the object was in rest)
- Height from the object falls = 2m.
- Time taken to fall = 1.08s.
To Find:-
- The Value of "g" i.e, Acceleration.
FORMULAE USED
- S = ut + ½ × a × t²
Where,
S = Distance
u = initial Velocity
a = Acceleration
t = Time
Now,
S = ut + ½ × a × t²
S = 0 × t + ½ × a × (1.08)²
S = 0 + ½ × 1.664 × a
S = 0.832 × a
2 = 0.832a
a = 2/0.832
a = 2.4m/s-²
Hence, The Value of "g" is 2.4m/s².
More Information
- The Value of g at the surface of the earth is 9.8m/s².
- The Value of "g" at the center of the earth is zero.
- The value of "g" is minimum in equator and maximum in Pole.
⭐ An Astronaut drops an object from a height of 2m and measures that it falls to the "ground" in 1.08 seconds. What is the value of "g" that can be calculated from this data?
✏ The value of "g" that can be calculated from the given data is 2.4 ms-².
- An Astronaut drops an object from a height of 2m and measures that it falls to the "ground" in 1.08 seconds.
- The value of "g" that can be calculated from the given data.
According to the question,
- An object is dropped from a height of 2m by an astronaut.
- It measures that it falls to the "ground" in 1.08 seconds.
We know ,
S = ut + ½ at²
Our given,
- Distance S = 2 meters
- Initial velocity u = 0
- Acceleration a = ?
- Time t = 1.08 seconds
Putting in the equation,
S = ut + ½ at²
=> 2 = 0 × 1.08 + ½ . a . (1.08)²
=> 2 = 0 + ½ . a. 1.17
=> a = 2 × 0.832
=> a = 2.4ms-²
=> a = 2.4 ms-²
The value of g by given data is 2.4 ms-² .
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