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EAMCET Engineering 2002 Solved Paper
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© examsnet.com
Question : 34
Total: 200
The horizontal and vertical displacements of a projectile at time
t
are
x
=
36
t
and
y
=
48
t
−
4.9
t
2
, respectively. Initial velocity of the projectile in
m
∕
s
is
15
30
45
60
Validate
Solution:
x
=
36
t
y
=
48
t
−
4.9
t
2
Comparing with
x
=
u
cos
θ
t
y
=
u
s
i
n
θ
t
−
1
2
g
t
2
where,
u
=
initial velocity
u
cos
θ
=
36
,
u
s
i
n
θ
=
48
(
u
cos
θ
)
2
+
(
u
s
i
n
θ
)
2
=
36
2
+
48
2
u
2
(
cos
2
θ
+
s
i
n
2
θ
)
=
3600
u
2
=
3600
u
=
60
m
∕
s
© examsnet.com
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