Question 1170979
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Since you provided no information on temperature, precipitation, air density, wind speed, or wind direction, I have to presume that this experiment was conducted in some sort of huge evacuated container and the effects of the atmosphere on the projectile are not to be considered.


The height in meters of a projectile near the surface of the earth at *[tex \Large t] seconds after it is launched is given by:


*[tex \LARGE (1) \ \ \ \ \ \ \ \ h(t) =\ -4.9t^2\ +\ \sin(\theta)v_ot\ =\ h_o]


Where *[tex \Large v_o] is the initial velocity in meters per second of the projectile, *[tex \Large \theta] is the angle of launch, and *[tex \Large h_o] is the initial height in meters. Then *[tex \Large \sin(\theta)v_o] is the vertical component of the initial velocity.


The instantaneous vertical velocity is given by the first derivative of the height function:


*[tex \LARGE (2) \ \ \ \ \ \ \ \ \frac{dh}{dt}\ =\ v(t)\ =\ -9.8t\ +\ \sin(\theta)v_o]


The maximum height of the projectile is the value of the height function at the time the velocity function is zero.  So set equation (2) equal to zero and solve for *[tex \Large t_{max}] and then calculate the value of *[tex \Large h\(t_{max}\)].


The projectile will be in the air for the amount of time it takes for the ball to reach zero height.  Set the height function equal to zero and then solve for the positive root which will be *[tex \Large t_{final}].


The final velocity will be the value of *[tex \Large v\(t_{final}\)]. 

<b>UPDATE</b>


The final vertical velocity will be the value of *[tex \Large v\(t_{final}\)].


The horizontal velocity is a constant *[tex \Large \cos(\theta\)v_o]


The final velocity magnitude is *[tex \Large \sqrt{v\(t_{final}\)^2\,+\,\(\cos(\theta\)v_o\)^2}]


The final velocity direction is *[tex \Large 180\ +\ \arctan\(\frac{v\(t_{final}\)}{\(\cos(\theta\)v_o}\)]
Make sure your calculator is set to degrees when you do the arctan calculation.


You can do your own arithmetic.


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