Q. A man throws a ball vertically upwards and it rises through 20 m and returns to his hands. What was the initial velocity $(u)$ of the ball and for how much time $(T)$ it remained in the air? $(g = 10\ ms^{- 2})$
A) $u = 10\ ms^{- 1};T = 2\ s$
B) $u = 10\ ms^{- 1};T = 4\ s$
C) $u = 20\ ms^{- 1};T = 2\ s$
D) $u = 20\ ms^{- 1};T = 4\ s$
Explanation
(d) $u = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = 20\ ms^{- 1}$
Time of ascent= time of descent
$= \sqrt{\frac{2h}{g} =}\sqrt{\frac{2 \times 20}{10}} = 2s$
$\therefore$ time of flight = 2 +2 = 4s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.