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Question

A projectile is given an initial velocity of (^i+2^j) m/s. The cartesian equation of its path is (g=10 m/s2)

A
y=2x5x2
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B
y=x5x2
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C
y=2x1.25x2
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D
y=2x25x2
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Solution

The correct option is A y=2x5x2
We have, u=^i+^2j
Comparing with u=ux^i+uy^j, we have
ux=1,uy=2
and tanθ=uyux=2
Now, ux=ucosθ,

Thus, the equation of the path of projectile is given by
y=xtanθgx22u2cos2θ
y=xtanθgx22(ux)2
y=x×210x22
y=2x5x2
Option A is correct.

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