A feather is dropped on the moon from a height of 2m. The acceleration of gravity on the moon is 1.69m/s sq. Determine the time for the feather to fall to the surface of the moon.
Answers
Given that, a feather is dropped on the moon from a height of 2 m. The acceleration of gravity on the moon is 1.69 m/s².
We have to find the time taken by the feather to fall to the surface of the moon.
From above data we have; height (s) is 2 m, acceleration due to gravity (a) is 1.69 m/s² and initial velocity (u) is 0 m/s.
First equation of motion:
v = u + at
Second equation of motion:
s = ut + 1/2 at²
Third equation of motion:
v² - u² = 2as
As we have value of of s, u and a. So, using the third equation of motion,
s = ut + 1/2 at²
Substitute the known values in the above formula,
→ 2 = (0)t + 1/2 × 1.69 × t²
→ 2 = 0 + 0.845t²
→ 2 = 0.845t²
→ 2.37 = t²
→ 1.54 = t
Therefore, time taken by the feather to fall to the surface of the moon is 1.54 sec.
Given:-
- Height of feather when that is dropped =2m
- Acceleration due to gravity on the moon = 1.69m/s²
- Initial velocity of feather = 0m/s
To Find:-
- The time taken by feather to fall on the surface of the moon.
Solution:-
we have to calculate the time taken by feathers to fall on the surface of moon.
so, By using 2nd equation of motion.
➦ s =ut + 1/2at²
2 = 0 × t + 1/2 × 1.69 × t²
2 = 1/2 × 1.69 × t²
2 = 0.845 × t²
t² =2/0.845
t² = 2.366 s
t = √2.366
t = 1.538 s