Question 1181323
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*[illustration VectorResolution.jpg].


The horizontal velocity *[tex \Large v_h] does not affect the height of the projectile.  The two vectors of interest when considering the height are the vertical acceleration vector which is equal to the acceleration due to the force of gravity, namely -9.8 meters per second squared (negative because the force is downward), and the vertical component of the initial velocity vector, namely *[tex \Large \|v_o_v\|\ =\ 85\sin30].  Presuming the missile is launched at ground level and assuming a missile ideal length of zero, the following quadratic models the instantaneous height at *[tex \Large t] seconds after launch:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ \frac{1}{2} \|a_v\|t^2\ +\ \|v_o_v\|t]


Divide the additive inverse of the first-degree term coefficient by the second-degree term coefficient to find the value of the independent variable at the vertex of the parabola.  The fact that the lead coefficient is negative ensures that this is the time value at the maximum height.


The maximum height reached is the value of the height function at the time calculated in the first part of the problem.


The horizontal distance is a function of the horizontal component of the initial velocity vector and the horizontal acceleration vector.  Since the horizontal acceleration vector has a magnitude of zero, the only relevant values are the total time of flight which is the non-zero root of the height function set to zero, and the horizontal component of the initial velocity.  The magnitude of the horizontal component of the initial velocity vector is *[tex \Large \|v_o_h\|\ =\ 85\cos30].  The horizontal distance traveled *[tex \Large s\(t_{max}\)] is then:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\(t_{max}\)\ =\ \|v_o_h\|t_{max}]


You can do your own arithmetic.

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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