no one tried it still? :/
is it too hard?

Let me drop some useful formula:

**Given:**
$$m=0.2\ \mathrm{kg}, h=10\ \mathrm{m}, b=0.4\ \mathrm{kg/s}, g=9.8\ \mathrm{m/s^2}$$

$$
k=\frac{b}{m}=\frac{0.4}{0.2}=2\ \text{s}^{-1}.
$$
Velocity:
$$
v(t)=\frac{g}{k}\big(1-e^{-kt}\big).
$$
Position (distance fallen after time (t)):
$$
y(t)=\int_0^t v(t')\ dt'=\frac{g}{k}t+\frac{g}{k^2}\big(e^{-kt}-1\big).
$$

$$
= \frac{g}{k}t+\frac{g}{k^2}\big(e^{-kt}-1\big)=h.
$$

now just plug in numbers...

noknees

science

Today's problem is a bit advanced 😅

**Problem:**
A 0.2 kg ball is dropped from rest from a height of **10 m**.
It experiences a linear drag force $$F_d = bv$$ with drag coefficient
$$
b = 0.4\ \text{kg/s}
$$

Find the **final velocity $$(v_f)$$** just before it hits the ground.

Answer should be correct to 2 decimal places with proper working **ONLY** required for finding $$t$$ and putting it into the velocity function.