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Question

A particle is projected with initial velocity v=(10^i+15^j)m/s in xy plane. The magnitude of displacement of the particle at time t=1 s is
(g=10^jm/s2)

A
10 m
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B
102 m
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C
20 m
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D
202 m
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Solution

The correct option is A 102 m
Given,
v=(10^i+15^j)m/s
g=10^jm/s2
t=1s
From 2nd equation of motion,
s=ut+12at2
s=(10^i+15^j)×1+12(10^j)(1)2
s=10^i+15^j5^j
s=10^i+10^j
Magnitude of the displacement,
|s|=(10)2+(10)2=200m
|s|=102m
The correct option is B.

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