How far does a baseball that is thrown horizontally at 42.5 m/s drop over a horizontal distance of 18.4 m?

Dec 6, 2015

$0.92 \text{m}$

Explanation:

The horizontal component of its velocity is constant so:

$t = \frac{s}{v}$

$\therefore t = \frac{18.4}{42.5} = 0.433 \text{s}$

For the vertical component:

$s = \frac{1}{2} \text{g"t^2=0.5xx9.8xx0.433^2=0.92"m}$