Question

(12.) 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​

Answer 1

Here in this case let us take that the mass of the ball is M.

Now the ball is thrown upwards with initial speed u but at the maximum height the ball stops and its speed becomes zero.

Now when the ball comes down then let us take that the ball hits the ground with speed V .

Now as the friction of air works on the opposite direction of the motion of the ball and the weight of the ball is w then ,

mass of the ball , M = w/g

Now when this ball comes downward then ,

$\frac{w}{g} \times a \: = (w - f) \\ a \: = \: \frac{(w - f)}{w} \times g$

where a is the acceleration of the ball in the downward direction.

Now from equation of motion we have,

$\implies$ V²=0+2ah

$\implies$V²=0+2ah⇒V²=2×(w−f)/w×g×h=2gh(w−f)/w

$\implies$V²=0+2ah⇒V²=2×(w−f)/w×g×h=2gh(w−f)/w⇒V=√2gh(w-f)/w

Hope it helps

Answer 2

Answer:

see the image

Explanation: Imagine the procedure of upward and downward pathway carefully and see the image .. hope it'll help