A ball of mass $2 k g$ and another of mass $4 k g$ are dropped together from a $60$ feet tall building. After a fall of $30$ feet each towards earth, their respective kinetic energies will be in the ratio of-

1 : 4

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1 : 2

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1 : \sqrt{2}

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\sqrt{2} : 1

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Solution

The correct option is B $1 : 2$

Here $V_{1} = V_{2} \sqrt{2 g h}$
$\frac{K E_{1}}{K E_{2}} = \frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} V_{2}^{2}} = \frac{m_{1}}{m_{2}} = \frac{2}{4} = \frac{1}{2}$

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Q. A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet, their respective kinetic energies will be in the ratio of

Q. A ball of mass

2 kg

and another of mass

4 kg

are dropped together from a

60

feet tall building. After a fall of

30

feet each towards earth, their respective kinetic energies will be in the ratio of -

Two balls of mass 2 kg and 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet, the ratio of their kinetic energy will be___. Neglect air resistance in both the cases.

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A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet each towards earth, their respective kinetic energies will be in the ratio of-

The correct option is B 1:2Here V1=V2√2ghKE1KE2=12m1v2112m2V22=m1m2=24=12

A ball of mass $2 k g$ and another of mass $4 k g$ are dropped together from a $60$ feet tall building. After a fall of $30$ feet each towards earth, their respective kinetic energies will be in the ratio of-

The correct option is B $1 : 2$

Here $V_{1} = V_{2} \sqrt{2 g h}$
$\frac{K E_{1}}{K E_{2}} = \frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} V_{2}^{2}} = \frac{m_{1}}{m_{2}} = \frac{2}{4} = \frac{1}{2}$