Question

A ball is thrown vertically upwards with speed u. If
it experiences a constant air resistance force of
magnitude f, then the speed with which ball strikes
the ground, (weight of ball is w) is
(1) v=uW+9"2
m
2
(w+f) 12
(3) v=u Tw-f)
(4) veu (w +f)) 12
(4) v = u Zw-F)​

Answer 1

Answer:

$u\sqrt{\frac{w-f}{w+f}}$

Explanation:

Weight of the ball $=w$

Mass of the ball $m=\frac{w}{g}$

The acceleration due to air resistance force $f$

$a=\frac{f}{m}$

$\implies a=\frac{fg}{w}$

When the ball is going up, the acceleration due to gravity will act downwards and the air resistance will also be downwards

$a_n=g+a$

When the ball is coming down, the acceleration due to gravity will act dowwards and the air resitance will act upwards

Therefore, the net acceleration

$a_n'=g-a$

Height covered by the ball while going up

using the third equation of motion

$v^2=u^2-2ah$

$0^2=u^2-2a_n\times h$

$\implies h=\frac{u^2}{2a_n}$

If the velocity with which the ball strikes the ground is v then again by third equation of motion

$v^2=0^2+2a_n'h$

$\implies v^2=2\times (g-a)\times \frac{u^2}{2(g+a)}$

$\implies v^2=u^2\times\frac{g-(fg/w)}{g+(fg/w)}$

$\implies v^2=u^2\times\frac{w-f}{w+f}$

$\implies v=u\sqrt{\frac{w-f}{w+f}}$

Hope this helps.