Hello!

I assume that we ignore air resistance and that "dropped" means zero initial velocity.

The only force which acts on either ball is the force of gravity. It acts downwards regardless of the position and the velocity of a ball. And we know that the gravity force for the ball with the mass `m` has the magnitude `mg.`

By Newton's Second Law the acceleration of a ball will be `(mg)/m=g` regardless of `m.`

Denote the height of a building as `H_0.` Both balls have the same constant acceleration `g,` the same initial height `H_0` and the same initial time (let it be zero).

Consider the projection to the upward axis. In this projection both balls also have the same initial velocity, it is zero (the projection of a horizontal initial velocity on the vertical axis is always zero). Therefore for both balls the height as a function of time will be

`H(t)=H_0-(g*t^2)/2.`

And the time `t_1` at which `H(t_1)=0` will be the same.

So the answer is **C, they land at the same time**.

P.S. The difference between the two balls movement is seen on the projection to the horizontal axis. For the first `L_1(t)=V_0*t,` for the second `L(t)=0` (L is for the length).

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