Question 1164086
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For a projectile launched vertically from near the earth's surface the function describing the height at time *[tex \Large t] is:


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


Where *[tex \Large g] is the acceleration due to the force of gravity which is *[tex \Large 9.8\text{ m/sec^2}] in the mks system, *[tex \Large v_o] is the initial velocity, and *[tex \Large h_o] is the initial height.


Plugging in the values you have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4.9(2.0)^2\ +\ v_o(2)\ +\ 1.5]


For part A, solve for *[tex \Large v_o]


For part B, instantaneous velocity is the value of the first derivative at any given instant, so:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dh}{dt}\ =\ -9.8t\ +\ v_o]


Insert the result of part A for *[tex \Large v_o] and then evaluate at *[tex \Large t\ =\ 2.0]

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