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UPSEE Solved Model Paper 2
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© examsnet.com
Question : 3
Total: 150
A bomber plane is moving horizontally with a speed of 500 m/s and a bomb released from it strikes the ground in 10 seconds. The angle with the horizontal at which the bomb strikes the ground is (g = 10 m/
s
2
)
t
a
n
−
1
(
1
)
t
a
n
−
1
(
5
)
t
a
n
−
1
(
1
5
)
s
i
n
−
1
(
1
5
)
Validate
Solution:
Let the vertical component of velocity when the bomb strikes the ground be
V
v
.
Let the horizontal component of velocity when the bomb strikes the ground be
V
H
.
V
v
= 0 + gt
Where, t = 10 seconds,
V
v
= 10 x 10 = 100 m/s and
V
H
= 500 m/s
tan θ =
V
v
V
H
=
100
500
=
1
5
θ =
t
a
n
−
1
(
1
5
)
© examsnet.com
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