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Test Index
TS EAMCET 5 May 2018 Shift 1 Solved Paper
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© examsnet.com
Question : 6
Total: 160
A small object is thrown at an angle 45° to the horizontal with an initial velocity
→
v
0
. The velocity is averaged for first
√
2
s , and the magnitude of average velocity comes out to be same as that of initial velocity i.e.
|
→
v
0
|
. The magnitude
|
→
v
0
|
will be
(Take g = 10 m/
s
2
)
3 m/s
3
√
2
m/s
4 m/s
5 m/s
Validate
Solution:
The displacement in x direction is
x
=
v
0
cos
θ
×
t
=
v
0
cos
45
°
×
√
2
=
v
0
×
1
√
2
×
√
2
=
v
0
The displacement along vertical direction is
y
=
v
0
sin
45
°
t
−
1
2
g
t
2
=
v
0
×
1
√
2
×
√
2
−
1
2
×
10
×
(
√
2
)
2
=
v
0
−
10
The total displacement is given by
d
=
√
v
0
2
+
(
v
0
−
5
)
2
The expression for average velocity is given as,
v
arg
=
√
v
0
2
+
(
v
0
−
5
)
2
√
2
√
2
v
0
=
√
v
0
2
+
(
v
0
−
5
)
2
2
v
0
2
=
v
0
2
+
(
v
0
−
5
)
2
v
0
=
5
m
∕
s
© examsnet.com
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