A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5\,m$, the number of ball thrown per minute is (take $g = 10\,m{s^{ - 2}}$)
A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5\,m$, the number of ball thrown per minute is (take $g = 10\,m{s^{ - 2}}$)
$120$
$80$
$60$
$40$
A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5\,m$, the number of ball thrown per minute is (take $g = 10\,m{s^{ - 2}}$)
Maximum height of ball $= 5 \,m$
So velocity of projection $ \Rightarrow u = \sqrt {2gh} $$ = 10\;m/s$
Time interval between two balls (time of ascent)
$ = \frac{u}{g} = 1\;sec = \frac{1}{{60}}\,min$.
So number of ball thrown per min. $= 60$
Other Language