a body is projected vertically upward with speed 40m/s. the distance travelled by body in the last second of upward journey is (take g=9.8m/s^2 and neglect effect of air resistance) (1)4.9m (2)9.8m (3)12.4m (4)19.6m
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s=ut+1/2at^2
u=0
a=g=9.8
at highest point t=1
s=0.5×9.8
s=4.9
u=0
a=g=9.8
at highest point t=1
s=0.5×9.8
s=4.9
Nani2818:
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I am Msc Physics student future professor.
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