When a object is shot from the bottom of a long smooth inclined plane kept at an angle 60 \degree with horizontal , it can travel a distance x_1 along the plane . But when the inclination is decreased  to 30 \degree and the same object is shot with the same velocity , it can travelx_2 \: \: distance .\: \: Then\: \: x_1 : x_ 2 \: \: will\: \: be :

  • Option 1)

    1 : \sqrt 2

  • Option 2)

    \sqrt 2 : 1

  • Option 3)

    1 : \sqrt 3

  • Option 4)

    1 : 2 \sqrt 3

Answers (1)

\Delta KE = - \Delta PE \\\\ \frac{1}{2} mV^2 = mgh \\\\ h_1 = h_2 \\\\ h_ 1 = x \sin 60 \\\\ h_ 2= x_ 2 \sin 30 \\\\ \frac{x_1 }{x_2 } = \frac{\sin 30 }{\sin 60}

= \frac{1}{2} \times \frac{1 }{\sqrt 3 /2 } = 1 / \sqrt 3


Option 1)

1 : \sqrt 2

Option 2)

\sqrt 2 : 1

Option 3)

1 : \sqrt 3

Option 4)

1 : 2 \sqrt 3

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