Question

A baseball is thrown straight up. the drag force is proportional to v2. part a in terms of g, what is the y-component of the ball's acceleration when its speed is half its terminal speed and it is moving up?

Answer 1

At the position of terminal speed the net acceleration of the ball will become zero

As we know that terminal speed will always reach when net force on the ball is zero and its speed will become constant.

So here at this position we can say

$F_d = F_g$

$kv^2 = mg$

$v =\sqrt{\frac{mg}{k}}$

now when ball is moving at half of the terminal speed in upward direction then net force on the ball in downwards direction will be

$F_{net} = mg + F_d$

$F_{net} = mg + kv^2$

here speed of the ball is half of the terminal speed

$v = \frac{1}{2}*\sqrt{\frac{mg}{k}}$

then we have

$F_{net} = mg + k*\frac{mg}{4k}$

$F_{net} = \frac{5mg}{4}$

now acceleration will be given as

$a = \frac{F_{net}}{m}$

now we have

$a = \frac{5g}{4}$

$a = 12.25 m/s^2$ downwards