Question

A ball falls freely from rest. The ratio of the distances travelled in first, second, third and fourth second is
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28.03.2019

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A ball falls freely from rest the ratio of the distance travelled in first , second , third and fourth sec is

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Brainly User

\huge{\mathfrak{Question:-}}

A ball falls freely from rest the ratio of the distance travelled in first , second , third and fourth sec is

\huge{\mathfrak{Solution:-}}

Distance travel is time t is:

\bold{s = ut + 1/2at^{2}}

Initial velocity u = 0

\bold{s = 1/2at^{2}}

Distance travel in 1 second = \bold{s_{1}= 1/2\times 9.8\times 1^{2}=4.9\;m}

similarly distance travelled in 2, 3, 4 seconds are:

\bold{s_{2}= 19.6\;m} \\ \\ \\ \bold{s_{3} = 44.1\;m} \\ \\ \\ \bold{s_{4} = 78.4\;m}

Now distance travelled in first (D1), second(D2), third(D3) and fourth(D4) second are:

\bold{D_{1}= s_{1}=4.9\;m}

\bold{D_{2}=s_{2}-s_{1}=19.6-4.9 =14.7\;m}

\bold{D_{3}=s_{3}-s_{2}=44.1-19.6 = 24.5\;m}

\bold{D_{4}=s_{4}-s_{3}=78.4 = 44.1= 34.3\;m}

\boxed{\sf{Ratio\;\;D_{1}:D_{2}:D_{3}:D_{4} = 1:3:5:7}}

I hope it will help you