A feather is dropped on the moon from a height of 2.52 meters. The acceleration of gravity on the moon is 1.67 m/s2. 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 2.52 m. The acceleration of gravity on the moon is 1.67 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 2.52 m, acceleration due to gravity (a) is 1.67 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,
→ 2.52 = (0)t + 1/2 × 1.67 × t²
→ 2.52 = 0 + 0.835t²
→ 2.52 = 0.835t²
→ 3.02 = t²
→ 1.74 = t
Therefore, time taken by the feather to fall to the surface of the moon is 1.74 sec.
Given :
▪ A feather is dropped from a height of 2.52 m.
▪ Acc. due to gravity of moon = 1.67 m/s²
To Find :
▪ Time taken by feather to reach at the surface of moon.
Concept :
☞ In this case, Constant acc. due to gravity continuously acts in downward direction, we can apply equation of kinematics to solve this question.
☞ For a free falling body, acceleration due to gravity is taken positive.
☞ Second equation of kinematics :
❍ s = ut + (1/2)at²
where,
s denotes distance
u denotes initial velocity
t denotes time
a denotes acceleration
Calculation :
For a free falling body, u = 0
→ s = (0×t) + (1/2)gt²
→ s = 0 + (1/2)gt²
→ t² = 2s/g
→ t = √(2s/g)
→ t = √[(2×2.52)/1.67]
→ t = √(5.04/1.67)
→ t = √3 s