A stone is dropped from a balloon when it is $150ft$ above the ground and rising at the rate of $10ft/sec$. How long will it take the stone to strike the ground, and with what velocity does it strike the ground?

I am not very familiar with antidifferentiation yet.

I think I should set

$v = speed$

$t = time (second)$

Would it be

$${dv \over dt} = 10$$

because it has a rising rate of 10ft/sec?

I am not sure where to start with this question.

  • 1
    $\begingroup$ do you mean $falling$ at a rate of $10 ft/sec$? $\endgroup$ – Joaquin Liniado Jun 23 '16 at 1:19
  • $\begingroup$ It might make more sense that way, but the question still states "rising at a rate of 10ft/sec" @JoaquinLiniado $\endgroup$ – didgocks Jun 23 '16 at 1:21
  • $\begingroup$ Is this a physics problem? Because gravity should definitely appear to make the stone fall... $\endgroup$ – Joaquin Liniado Jun 23 '16 at 1:23
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    $\begingroup$ The balloon is the one rising $\endgroup$ – Deepak Jun 23 '16 at 1:26
  • $\begingroup$ This is from calculus book, perhaps I should skip this question? $\endgroup$ – didgocks Jun 23 '16 at 1:27

Setting coordinates so that positive velocities are downwards, you have $\frac{dv}{dt}$ is the acceleration due to gravity, $g$. So you want to solve $\frac{d^2s}{dt^2} = \frac{dv}{dt}$ subject to two initial conditions. Can you extract the initial conditions from the data you are given?

  • $\begingroup$ This is all the conditions given from the question, but the answer is $5.9 sec$, and $188 ft/sec$ $\endgroup$ – didgocks Jun 23 '16 at 1:32

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