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Question

A body is thrown vertically up with a velocity u.if it passes 3 points A,B,C in its upward journey with the velocity U/2, U/3, U/4 respectively. The ratio of separation AB and BC is?

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Solution

Step 1: Given

Initial velocity of the body = u.

Velocity at point A, uA=U2

Velocity at point B, uB=U3

Velocity at point C, uC=U4


Step 2: Formula used and calculation

To find distance AB, use third equation of motion, v2=u2+2as

Here, u=U2,a=g,v=U3

(U3)2=(U2)2+2(g)sAB

sAB=5U272g

Similarly to find distance BC, use third equation of motion

Here, u=U3,a=g,v=U4

(U4)2=(U3)2+2(g)sBC

sBC=7U2288g

Now,

sABSBC=5U272g7U2288g

sABSBC=207


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