# Question bbfa9

Jul 21, 2015

Time: 3.57 s

#### Explanation:

When you fire a projectile horizontally from a certain height, the only parameter that will influence how much time that projectile spends in flight is the height from which you launched it.

In other words, the time of flight depends solely on the height of the pirate ship. The initial velocity will impact the projectile's range, but not its flight time.

Vertically, the movement of the projectile will be affected by gravity. Since you start at a height of 17.5 m above sea level and enp up at sea level, the vertical displacement of the projectile will be negative.

Don't forget to use this convention for the sign of the gravitational acceleration as well!

This means that you can write

$- h = {\underbrace{{v}_{0 y}}}_{\textcolor{b l u e}{\text{=0}}} \cdot t - \frac{1}{2} g \cdot {t}^{2}$

The vertical component of the initial speed will be zero because the projectile is launched parallel to the horizontal axis.

${v}_{0 y} = {v}_{0} \cdot \sin \left({0}^{\circ}\right) = 0$

As a result, the time of flight will be

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

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

t = sqrt((2 * 17.5cancel("m"))/(9.8 cancel("m")/"s"^(2))) = color(green)("3.57 s")#