. Explain Motion of a particle projected vertically upward from the ground
i. Consider a particle projected vertically upward from the ground with velocity u.
ii. Taking upward direction positive u = u, a = – g. At any time t, velocity v = u – g t and displacement s= ut-1/2gt2
iii. To find time of ascent(ts) , apply v = u + at between the point of projection and the highest point.
v = 0, u = u, a = – g.
vi. Time of descent (td) between the highest point and back to the point of projection is also u/g : td= u/g
vii. The particle will return back to the point of projection with same speed as the speed of projection but in the opposite direction.
viii. Motion under gravity is symmetric.