The equation(adsbygoogle = window.adsbygoogle || []).push({});

v_{2}^{2}= v_{o}^{2}+ 2aΔy

is used to determine the change in position, acceleration, or velocity of a particle under the influence of a quadratic force.

on earth, the acceleration value in this equation would, under normal circumstances, be

a= -9.8 m/s^{2}.

so, this substitutes into

v_{2}^{2}= v_{o}^{2}+ -19.6Δy

In this

Δy = v_{2}^{2}- v_{o}^{2}/ -19.6

Lets now assume that v_{2}^{2}is 0, the crest of a projectiles motion.

Δy = 0 - v_{o}^{2}/ -19.6

If the objects initial v is facing upwards, then obviously the net Δy would be positive.

Now, keeping all conditions the same, lets assume that the acceleration is 9.8m/s^{s}upwards

This would yield

Δy = 0 - v_{o}^{2}/ 19.6

Δy would thus be negative, and the object would be thrown downwards, accelerating upwards.

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# Basic Kinematics Question

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