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The question tells me that I throw a ball directly upwards with a mass = m, initial velocity = U, the velocity at any point = V, air resistance = kv^2 and W = terminal velocity. I have to find the time at max height in terms of V, the max height it reaches and an equation incorporating terminal velocity as it descends.

I started with $a = g - k\frac {v^2}{m}$

Using $a = v\frac{dv}{dx}$, i found that $x_{max} = \frac {m}{2k} * ln\bigl (1-\frac {kU^2}{mg}\bigr).$

However, I don't know how to find the time at max height. I assume I go back to the original equation for a and use $a = \frac {dv}{dt}$ as such: $dt = \frac {m}{mg-kv^2}*dv$.

I am unsure how to best integrate this. I don't think trig substitution works as I believe it results in a $tan(x)$ and the angle is 90 degrees. Do I attempt partial fractions? Or else how do I approach integrating it?

Furthermore, I don't understand how to calculate terminal velocity. Any tips would be greatly appreciated!

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    $\begingroup$ I still this fits better in SE physics. $\endgroup$ – Evargalo Aug 10 '17 at 7:41
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    $\begingroup$ @Evargalo It is a question from my Maths C class, so I figured it belonged in SE maths $\endgroup$ – greenie Aug 10 '17 at 7:43
  • $\begingroup$ physics are maths btw $\endgroup$ – Thomas Prévost Aug 10 '17 at 8:12
  • $\begingroup$ isn't it $a = -g-k\frac {v^2}{m}$ ? $\endgroup$ – Thomas Prévost Aug 10 '17 at 8:28
  • $\begingroup$ I have been taught to always have g = -9.8. $\endgroup$ – greenie Aug 10 '17 at 8:30
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HINTS...To obtain the maximum height, your equation of motion should be $$ma=-mg-kv^2.$$ The expression you have is for the downward motion, not the upward motion.

Use $a=\frac{dv}{dt}$ and get an arctan integral to find the time at max height

To get the terminal velocity put $a=0$ In the downward equation.

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