When you drop a ball it accelerates downward at 9.8 m/s^2. If instead you 5 points throw it downwards, then its accleration immediately after leaving your
hand is?

if you need an actual number answer you can use   :Vf = Vi + at.  If you throw the ball it will have an initial force beside gravity accelerating the fall temporarily from greater than throw it downwards, its acceleration (in the absence of air resistance) will be greater than  9.8 m/s2 until it slows back down to a constant 9.8 m/s2 after ( t )amount of time

Explanation:

If you drop a ball, it accelerates downward at 9.8 m/s2. if instead you throw the ball straight downwards While throwing, we apply an additional force other than the gravitational force.

This gives an additional, temporary acceleration along with the gravitational acceleration.

Thus from the instant it is thrown and the instant it leaves your hand, the object is under variable acceleration, the variation of acceleration being the reason of the varying force which we do apply on the object. But once it leaves our hand it is always under constant acceleration of g which is9.8 m/s2

explanation:

newton's third law, for every action there is an equal and opposite reaction.

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