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well, assuming no outside interference such as wind, air resistance, etc...

the acceleration is -9.8m/s squared.

the time is 2 seconds

and the initial velocity in the vertical plane is 0

so, using the formula d=v(1)t+1/2at(squared)

d=1/2(-9.8)(2)squared

d= -19.6m

so, 19.6m beneath the straight line path

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Q: How far beneath the straight line path does the projectile fall after 2 seconds?
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