# Does horizontal inertia affect the time it takes something to reach the ground?

I am new to physics.

I’m confused about how an bullet shot horizontally would land at the same time with as a bullet dropped vertically (ignoring air resistance and the curvature of Earth).

Having just learnt about inertia previously, I am confused: Why wouldn't it take time for gravity to overcome the horizontal inertia of the bullet?

Maybe I am confusing momentum with inertia? Does inertia increase with velocity? (I haven't learnt that yet)

EDIT: I realize my mistake now. I saw a video on inertia that explained how cars must first overcome their "forwards inertia" in order to turn. This misled me into believing horizontal/vertical inertia were dependent.

If you shoot a bullet horizontally with the velocity $v_x$, the gravitational force acts perpendicular to the initial trajectory of the bullet. Since you can decompose a particle's motion (in particular its velocity) into its orthogonal components, you have a component $v_x$ for the horizontal motion and a component $v_y$ for its vertical motion (i.e. towards the ground). The gravitational force leads to an acceleration of the bullet towards the ground, and the acceleration is always in the same direction as the force. Therefore, gravitation will only change the vertical speed component $v_y$. More important for your question, gravitation does not "care" whether there is an initial horizontal velocity. $\vec{F} = \begin{pmatrix} 0 \\ -mg \end{pmatrix} = m\cdot \vec{a}$.
Note that only the y component of $\vec{a}$ is non-vanishing. If you want to calculate the time it takes for the bullet to hit the ground, you need the formula $s_y = \frac{1}{2} a_y t^2$, assuming that no initial velocity in the direction of $y$ is given. In the direction of $x$, the bullet is not accelerated, i.e. $a_x = 0$. Solving for $t$ leads to: $t = \sqrt{\frac{2s}{a_y}}$. Hence, the time is independent from $v_x$.