an object is thrown vertically upwards with a velocity of 19.6m/s. calculate the distance and displacement of the object after 3 seconds. ans: distance=24.5 displacement=14.7
u=19.6m/s
t=3secs.
s=ut +1/2at2
s=19.6(3) +1/2(-9.8)(9)
=58.8-44.1
=14.7m
displacement=14.7m
v=u-gt
v=0
u=19.6m/s
0=19.6-9.8t
for going up,
t=2secs
s=19.6(2) + 1/2(-9.8)(2)2
=19.6m is the max height attained.
for coming down,
u=0
t= 3 secs- time taken to go up
t=3-2
=1 sec
s=0+1/2(9.8)(1)
=4.9m
total distance=distance gone up + distance come down
=19.6 + 4.9
=24.5m
therefore,
total distance=24.5m