A physics student falls from 12.0 cliff How long does it take him to reach the ground and what is his velocity by the time he reaches the ground?
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Answer:
Height of the cliff h
1
=20m
Velocity of projectile V=30m/s
Initial upward velocity v
y
=Vsinθ=30×sin30
∘
=15m/s.
Horizontal velocity, v
x
=Vcosθ=30cos30
o
m/s
Time taken to reach maximum height, t
1
=
g
V
y
=
10
15
=1.5sec.
Height above the cliff h
2
=
2g
v
y
2
=
2×10
15
2
=11.25m
Therefore, the maximum height, h
max
=h
1
+h
2
=20+11.25=31.25m.
Time taken to free fall from max height, t
2
=
g
2h
max
=
g
2×31.25
=2.5sec.
Thus, the total time taken during the entire flight t
total
=t
1
+t
2
=1.5+2.5=4sec
Total horizontal distance covered R=v
x
t
total
=30×cos30
∘
×4=60
3
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