If a ball is thrown vertically with air resistance = kv^2, how long till max height?

Question

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!

@Evargalo It is a question from my Maths C class, so I figured it belonged in SE maths — greenie, Aug 10 '17 at 07:43
ah ok, it's because I use $g = 9.8$ and your ball goes against it... so I used $-g$... — Thomas Prévost, Aug 10 '17 at 08:33
Definitely a maths question. To solve $$t=\int_{v(t)}^u\frac{m,dz}{kz^2-mg}$$ note that $$\frac{2a}{z^2-a^2}=\frac1{z-a}-\frac1{z+a}$$ hence, for $a=\sqrt{mg/k}$, $$\frac{m}{kz^2-mg}=\frac{a^2/g}{z^2-a^2}=\frac{a}g\left(\frac1{z-a}-\frac1{z+a}\right)$$ which yields $$\int\frac{m,dz}{kz^2-mg}=\sqrt{\frac{m}{kg}}\log\left|\frac{z-a}{z+a}\right|+C$$ Can you continue? — Did, Aug 10 '17 at 08:35

score 1 · Accepted Answer · answered Aug 10 '17 at 08:34

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.

If a ball is thrown vertically with air resistance = kv^2, how long till max height?

1 Answers1