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?

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Answer 1 · 2018-03-07T22:19:34

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

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

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

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

$h = v_{y} \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $h = v_y*t + 1/2*g*t^2$

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

$h = v_{y} \cdot t + \frac{1}{2} \cdot g \cdot t^{2} = \frac{1}{2} \cdot g \cdot t^{2}$ $h = v_y*t + 1/2*g*t^2 = 1/2*g*t^2$

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