Question 1168960
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The height, *[tex \Large h], as a function of time, *[tex \Large t], neglective air resistance for a projectile near the surface of the earth (gravitational acceleration in the MKS system *[tex \Large g\,=\,-9.8\text{ m/s^2}])with a vertical component of initial velocity equal to  *[tex \Large v_o] and an initial height equal to *[tex \Large h_o] is:


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


Note that *[tex \Large g] is a negative value because gravitational acceleration is toward the center of the massive body.


The extreme value of this function occurs at *[tex \Large t_{max}] which is the independent variable value that makes the first derivative of the function equal to zero.  This extreme value is a maximum if the second derivative is negative at that point.



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dh}{dt}\ =\ gt\ +\ v_o]


So solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ gt_{max}\ +\ v_o]


for *[tex \Large t_{max}] noting that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2h}{dt^2}\ =\ g]


which is a constant *[tex \Large g\,=\,-9.8\text{ m/s^2}] so the extremum is a maximum.


The time that the projectile is in the air is the positive zero of the function.  So set the function equal to zero and solve the quadratic.  Of course, your difficulty is that you don't know how high off the ground the ball was when the person throwing it released it.  So you only have a very vague idea of what the value of *[tex \Large h_o] might be.


Good luck.

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


I > Ø
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