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Question

The equation of projectile is y=8x5x22. The horizontal range is-

A
8 m
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B
3.2 m
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C
8.4 m
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D
2 m
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Solution

The correct option is B 3.2 m
Equation of projectile is given by y=x tan θ(1xR)
y=x tan θtan θx2R
Here R is Range of projectile.

Given that y=8x5x22
Compare from eqn.
tan θ=8 and tan θR=52
R=165=3.2 m
R=3.2 m

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