Question

particle is thrown vertically upwards with velocity 40 m/s after some time particle will return to the point of projection assuming upward direction positive, sketch velocity -time graph, displacement - time graph and acceleration - time graph.

Answer 1

in velocity time graph--
initial velocity= 40m/s
final velocity=0
so , time taken =
by 1st eq. --
v=u+at
t= 4 sec to reach the highest point
total time of journey = 2t = 8 sec.

2). in displacement time graph--
final velocity is equal to zero. during upward motion.
an initial velocity is 40 m per second
so by third equation of motion--
v²=u²+2as
a= g = 10m/s² ( downward direction)
so, s = 80m (during our journey)

in acceleration time graph--
in all the journey acceleration will remain constant , only one acceleration is acting upon the body is G= 10 m/s²
now , graphs-- in the attachment..
hope it helps..