# A soccer ball is kicked horizontally at 18.8 m/s off the top of a building and lands 33.9 meters from the building. What is the height of the building.? (round answer to 3 sig. figs) Answer: ______ m

Nov 3, 2016

$15.9 m$, rounded to 3 sig. figs

#### Explanation:

The soccer ball travels a horizontal distance of $33.9 m$ after being kicked with horizontal velocity of $18.8 m {s}^{-} 1$. Assuming air resistance is negligible.

Time of flight of the ball $t = \text{Distance"/"Speed} = \frac{33.9}{18.8} s$

In the vertical direction kinematic equation is
height $h = u t + \frac{1}{2} g {t}^{2}$
Initial velocity ${u}_{v} = 0$, acceleration due to gravity $g = 9.81 m {s}^{-} 2$. Inserting various values we get
$h = 0 \times \frac{33.9}{18.8} + \frac{1}{2} \times 9.81 \times {\left(\frac{33.9}{18.8}\right)}^{2}$
$\implies h = \frac{1}{2} \times 9.81 \times {\left(\frac{33.9}{18.8}\right)}^{2}$
$\implies h = 15.9 m$, rounded to 3 sig. figs