# Terminal speed

will a golf ball of mass 45g and diamter 4.3cm reach terminal speed when dropped from a height of 25m? the drag coefficient is 0.35 and the density of air is 1.2kg/m^3.

using the formula $$v_t = sqrt(\frac{2*mg}{CpA})$$

$$v_t = sqrt(\frac{2*(45g)(9.8m/s^2)}{(.35)(1.2kg/m^3)(0.043m)})$$

so pluggeed that into my calculator, i found the terminal speed of 220.99m/s.
this is where i got stuck, how would i know if it reaches terminal speed or not when dropping from a height of 25m?

it will not reach the terminal speed nomatter how high you drop the ball......
the velocity of the ball is......
v=v_t(1-e^(-kt))
it will get closer and closer to the terminal speed when time passes, but it will never "reach" it

i have a few questions, how did you get the formula v=v_t(1-e^(-kt))? and what does k and t stand for and how do i find it?

solve the following DE
$$-m\frac{dv}{dt}= -mg+ bv$$
b is the drag coefficient (i believe this is how ppl called it)
k=b/m ...and t is time...

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sorry but a few more questions if you dont mind. how extactly would i find t? I dont think using one of the kenematics will help because i will be missing alot of info.

you know calculus, do you?
$$m\frac{dv}{dt}= mg- bv$$
$$\frac{dv}{dt}=g-\frac{b}{m}v$$
$$\frac{dv}{dt}=g-kv$$
$$dt = \frac{dv}{g-kv}$$
$$\int dt = \int \frac{dv}{g-kv}$$
$$t = \frac{-1}{k} ln(g-kv) + C$$
$$v(t) = \frac{g}{k}-\frac{e^{-k(t-C)}}{k}$$
$$v(t) = \frac{g}{k}-\frac{C'e^{-k(t)}}{k}$$

apply the initial condition $v(0)=0$ this implies $C'=g$
therefore,
$$v(t) = \frac{g}{k}-\frac{ge^{-k(t)}}{k}$$
$$v(t) = \frac{g}{k}(1-e^{-k(t)})$$
$$v(t) = \frac{gm}{b}(1-e^{-mt/b})$$

whereas gm/b is the terminal velocity

ah i see, thanks for the help!