Question

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?

Answer 1

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.

Answer 2

$\mathcal{\huge{\fbox{\fbox{\red{Question\:?}}}}}$

⭐ 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?

$\mathcal{\huge{\fbox{\fbox{\green{Answer:-}}}}}$

✏ The value of "g" that can be calculated from the given data is 2.4 ms-².

$\mathcal{\huge{\fbox{\fbox{\purple{Solution:-}}}}}$

$\mathtt{\huge{\underline{Given:-}}}$

• An Astronaut drops an object from a height of 2m and measures that it falls to the "ground" in 1.08 seconds.

$\mathtt{\huge{\underline{To\:Find:-}}}$

• The value of "g" that can be calculated from the given data.

$\mathtt{\huge{\underline{Calculation:-}}}$

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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