# Antidifferentiation: Stone dropped from $150ft$ rising at $10ft/sec$

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?

• do you mean $falling$ at a rate of $10 ft/sec$? – Joaquin Liniado Jun 23 '16 at 1:19
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?
• This is all the conditions given from the question, but the answer is $5.9 sec$, and $188 ft/sec$ – didgocks Jun 23 '16 at 1:32