Question

Refer to the attachment .


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Answer 1

Given,  acceleration a=1.67m/s2 

Distance covered by feather d=1.40m 

Initial velocity u=0m/s (assume heather falls from rest)

Using formula S=ut+at2/2,

1.40=0(t)+21(1.67)t2

or t2=2.8/1.67=1.68 

⟹ t=1.68=1.29s

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Answer 2

Answer:

• 1.29 s.

Explanation:

Given :

$\Rightarrow \bold{Height\;(s)=1.4 m}$

$\Rightarrow \bold{Acceleration\;due\;to\;gravity\;on\;moon(a)=1.67m/s^2}$

$\Rightarrow \bold{Initial\;velocity\;of\;the\;feather(u) = 0m/s}$

To find :

$\Rightarrow \bold{Time\taken\;for\;the\;feather\;to\;fall\;on\;surface(t)}$

Solution :

Using kinematics:

${\boxed{\bold{s=ut+\frac{1}{2}at^2} }$

Substituting:

$\rightarrow\bold{1.4=0(t)+\frac{1}{2}(1.67)t^2 }$

$\rightarrow \bold{t^2=\dfrac{2.8}{1.67} }$

$\rightarrow\bold{t^2=1.67}$

$\rightarrow\bold{t=\sqrt{1.67}=1.29 s }$

∴ The time taken for the feather to the fall on the surface is 1.29 s.