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anonymous

  • one year ago

A hot-air balloon is ascending at a rate of 7.5m/s when a passenger drops a camera. If the camera is 25m above when it is dropped, (a) how much time does it take for the camera to reach the ground, and (b) what is its velocity just before it lands. Upward it the positive direction in this problem.

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  1. anonymous
    • one year ago
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    |dw:1444116101171:dw|

  2. anonymous
    • one year ago
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    see as every sec. the balloon is ascending up 7.5 m . thus each particle , each object inside the balloon is also moving with the same vel. of 7.5 m/s... so when something separate with the balloon like camera in this case would also have same vel. thus initial vel. of camera = 7.5 m/s [ as vel in upward direction ] acc. due to gravity = - 9.8m/s sq.\[s=ut + \frac{ 1 }{ 2 }at^2\]

  3. anonymous
    • one year ago
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    -25 = 7.5 * t - 4.9*t^2 [as dist. to be covered is in -ve direction] 4.9t^2 -7.5t - 25 =0

  4. anonymous
    • one year ago
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    \[v^2-u^2 =2as\] to find the vel just before hitting the ground where u=+ve , a=-ve , s=-ve

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