Basic Kinematics Question

The equation


v₂² = v_o² + 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/s².


so, this substitutes into

v₂² = v_o² + -19.6Δy

In this

Δy = v₂² - v_o² / -19.6

Lets now assume that v₂² is 0, the crest of a projectiles motion.

Δy = 0 - v_o² / -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_o² / 19.6

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