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1st Equation of Motion

A freely fall...

Question

A freely falling body crosses points P, Q and R with velocity V, $2$ V and $3$ V respectively. Find the ratio of the distance PQ to QR.

5 : 3

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3 : 5

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1 : 2

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2 : 1

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Solution

The correct option is B $3 : 5$
Let the acceleration be $a$ .
For P to Q :
Initial speed $u = V$
Final speed $v = 2 V$
Using $v^{2} - u^{2} = 2 a S$
$∴$ $2 a S_{P Q} = (2 V)^{2} - V^{2}$
We get $S_{P Q} = \frac{3 V^{2}}{2 a}$
For Q to R :
Initial speed $u = 2 V$
Final speed $v = 3 V$
Using $v^{2} - u^{2} = 2 a S$
$∴$ $2 a S_{Q R} = (3 V)^{2} - (2 V)^{2}$
We get $S_{Q R} = \frac{5 V^{2}}{2 a}$
So. $S_{Q R} : S_{P Q} = 3 : 5$

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A freely falling body crosses points P, Q and R with velocity V, 2V and 3V respectively. Find the ratio of the distance PQ to QR.

The correct option is B 3:5Let the acceleration be a.For P to Q :Initial speed u=VFinal speed v=2VUsing v2−u2=2aS∴ 2aSPQ=(2V)2−V2We get SPQ=3V22aFor Q to R :Initial speed u=2VFinal speed v=3VUsing v2−u2=2aS∴ 2aSQR=(3V)2−(2V)2We get SQR=5V22aSo. SQR:SPQ=3:5

A freely falling body crosses points P, Q and R with velocity V, $2$ V and $3$ V respectively. Find the ratio of the distance PQ to QR.