# Question 564a5

Jun 13, 2015

The time of flight is $t = \text{10 s}$.
The horizontal component of its displacement is ${d}_{x} = \text{1500 m}$.

#### Explanation:

Here's what's going on after the object is dropped from the plane.

There are two distinct directions for which you must describe the motion of the object.

Horizontally, the initial velocity of the object will be equal to that of the plane. Moreover, since no horizontal forces act upon the object, its horizontal motion takes place at constant speed.

This means that the horizontal component of its displacment can be written as

${d}_{x} = {v}_{0 x} \cdot t$, where

${v}_{0 x}$ - the horizontal component of the initial velocity;
$t$ - the total time of the fall.

Vertically, falls under the influence of gravity. This means that its vertical motion will be accelerated, the actual acceleration being equal to the gravitational acceleration, $g$.

The object starts at the height at which the plance cruises, and its initial velocity has a Zero vertical component. Moreover, it drops down in the same total time $t$. This means that you can write

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

${d}_{y} = \frac{1}{2} g {t}^{2}$

Plug in your data and solve for $t$

$500 \frac{\cancel{{\text{m") = 1/2 * 10cancel("m")/"s"^2 * t^2 => t^2 = (500"s}}^{2}}}{5}$

$t = \sqrt{100 \text{s"^2) = color(green)("10 s}}$

The object drops for a total time of 10 seconds. This means that the horizontal component of its displacement will be

d_x = 150"m"/cancel("s") * 10cancel("s") = color(green)("1500 m")#