Question 485107
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Your second problem:


You need to choose the measurement system you want to use.  Acceleration due to gravity in kms is *[tex \Large g\ =\ -9.8 m/sec^2], whereas in fps it is *[tex \Large g\ =\ -32 ft/sec^2].  The values are negative because of the convention that up is positive.


I'm just going to do the problem using *[tex \Large g] and then you can substitute the appropriate acceleration value and do the arithmetic.


The instantaneous velocity of a falling body is the integral of acceleration with respect to time plus a constant which is the initial velocity, *[tex \Large v_o]


Since the rock was dropped we can presume that *[tex \Large v_o\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v(t)\ =\ \int\,g\ dt\ =\ gt\ +\ v_o]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v(t)\ =\ gt]


Substitute 2.5 seconds for the time because that is what is given and multiply by the appropriate value of *[tex \Large g].  If you use *[tex \Large -9.8] your answer will be in meters per second, but if you use *[tex \Large -32] then your answer will be in feet per second.


The height is determined by integrating velocity as a function of time.  Very conveniently the function we just evaluated to find the final velocity.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,gt\ +\ v_o\,dt\ =\ \frac{1}{2}gt^2\ +\ v_ot\ +\ C]


Where the constant of integration is the initial height, *[tex \Large h_o]


For simplicity's sake, we will take the initial height to be zero and then consider the surface of the water to be some negative height.


Again, we substitute 0 for the initial values and get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -\frac{1}{2}gt^2]


And again, substitute 2.5 for *[tex \Large t] and your chosen value for *[tex \Large g]





John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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