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Question

A particle is released from a certain height of 600 m. Due to the wind, the particle gathers a horizontal velocity component of ay where a is 10 s1 and y is the vertical displacement of the particle from the point of release. What is the horizontal drift of the particle when it strikes the ground?
[Take g=10 m/s2]

A
7315 m
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B
7215 m
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C
7115 m
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D
7015 m
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Solution

The correct option is D 7015 m
Given: Height, H=600 m
Horizontal velocity gained due to wind,
vx=ay=10y at distance y below point of release
The time of flight of the particle,
t=2Hg=2×6001011 s
Now, vx=10y
dxdt=10y
x0dx=10t0y dt
x0dx=10t012gt2 dt
x0dx=510110t2 dt
x=510[t33]110
x=510[(11)330]
x7015 m
Hence, the option (d) is correct.

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