JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 7)
A Tennis ball is released from a height h and after freely falling on a wooden floor
it rebounds and reaches height $${h \over 2}$$. The
velocity versus height of the ball during its motion may be represented graphically by :
(graph are drawn schematically and on not to scale)
(graph are drawn schematically and on not to scale)
_4th_September_Morning_Slot_en_7_1.png)
_4th_September_Morning_Slot_en_7_2.png)
_4th_September_Morning_Slot_en_7_3.png)
_4th_September_Morning_Slot_en_7_4.png)
توضیح
V, h curve will be parabolic.
Downward velocity is negative and upward is positive
When ball is coming down graph will be in IV quadrant and when going up graph will be in I quadrant.
At H = h, v = 0
at h = 0, v = $$\sqrt {2gh} $$
also a = –g, throughout this motion.
Downward velocity is negative and upward is positive
When ball is coming down graph will be in IV quadrant and when going up graph will be in I quadrant.
At H = h, v = 0
at h = 0, v = $$\sqrt {2gh} $$
also a = –g, throughout this motion.