In projectile motion, why does an object that is vertically dropped down and an object that is horizontally launched reach the ground at the same time?

Mar 7, 2018

Because acceleration due to gravity is $9.8 \frac{m}{s} ^ 2$ regardless of any horizontal velocity.

Explanation:

A dropped object accelerates downward at g. The time to the ground is given by

$h = \frac{1}{2} \cdot g \cdot {t}^{2}$

A projectile launched horizontally accelerates downward at g. The time to the ground is given by

$h = {v}_{y} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

where the vertical component of its velocity, ${v}_{y}$, is zero. Therefore

$h = {v}_{y} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2} = \frac{1}{2} \cdot g \cdot {t}^{2}$

One condition: if the ground is not perfectly level from the site of the launch of the projectile to the impact area, that would change things.

I hope this helps,
Steve