Physics, asked by devanshagarwal013, 20 days ago

a stone is dropped from a rising balloon at a height of 470.4 m. if the balloon is rising at a speed of 19.6m/s, when and with what velocity does the stone hit the ground.
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Answers

Answered by Anonymous
2

The balloon is rising at in upward direction of motion = 19.6 m/s¹. So, u = –19.6 m/s¹.

When the loon reached at the height of 470.4 m, it dropped the stone from that height.

Now, The body ( stone ) is projected vertically downward with some initial velocity from some height.

Which means the stone is falling under gravity. So, by using this formula of kinematics, we get:

  • v² = u² + 2gh
  • v² = ( –19.6 )² + 2 × 9.8 × 470.4
  • v² = 384.16 + 9219.84
  • v² = 9604
  • v = √9604 = 98 m/s¹. ( It's the required velocity of stone )

Now, using this formula we get:–

  • v = u + at
  • 98 = ( –19.6 ) + ( 9.8 × t )
  • 98 + 19.6 = 9.8t
  • 9.8t = 117.6
  • t = 12s. ( It the total time taken by that stone to reach at the ground from the height of 470.4 m )

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