Question

a ball is thrown vertically upwards with speed u. If its experiences a constant air resistance force 'f' then speed with which ball strikes the ground is (w is weight of ball)
the answer is of the form v = u [(w-f)/(w+f)]^1/2

Answer 1

Dear student

When ball is thrown vertically upward with velocity u. The weight w and air resistance force work downward so let the acceleration during motion is a and height covered by the ball is h then
$$\begin{gathered} 0=u^2-2ahor \\ {u}^2=2ahandbyNewton\text{'}ssecondlaw \\ \left(w+f\right)=ma \\ wherelemisthemassoftheballthen \\ \left(w+f\right)=\frac{u^2}{2h}.......\left(1\right) \end{gathered}$$
Then ball fall from rest from height h and let the speed of the ball is v when it hit the ground. Now the air resistance is working on the ball upwards so net force on the ball is (w-f) and let the acceleration is a' so
$$\begin{gathered} v^2=0+2a\text{'}hand \\ \left(w-f\right)=ma\text{'} \\ \left(w-f\right)=\frac{m{v}^2}{2h}.......\left(2\right) \\ Nowdividetheeq./\left(1\right) \\ \frac{\left(w-f\right)}{\left(w+f\right)}=\frac{\frac{m{v}^2}{2h}}{\frac{m{u}^2}{2h}} \\ \frac{\left(w-f\right)}{\left(w+f\right)}=\frac{v^2}{u^2} \\ sov=u{\left(\frac{w-f}{w+f}\right)}^{1/2} \end{gathered}$$