A bullet is fired vertically upwards with an initial velocity of 50m/s.it covers a distance h1 during the first second and a distance h2 during the Last 3 seconds of its upward motion.if g=10m/s^2.what is the relation between h1 and h2?
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Answer:
using equation of motion :s=ut+
2
at
2
distance traveled in first second:
h
1
=45m
t
a
=
g
u
,where t
a
is time of ascent
t
a
=5s
so at t=2s last 3 sec of upward motion starts
height at t=2s is :
h=50×2−
2
10×4
h=80m
H
max
=
2g
u
2
Hmax=125m
distance traveled in last 3 sec h
2
=H−h
h
2
=125−80
h
2
=45m
therefore h
1
=h
2
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