P02R2015 – Air Delivery

First, we setup the kinematic equations of the radar decoy. Since it falls solely under its own weight, its vertical acceleration is $a_y=-g$ while its horizontal acceleration is $a_x=0$.

The components of the initial velocity of the decoy are $\displaystyle{\left\{ \begin{array}{c}
v_{0x}=v_0{\mathrm{cos} \left(\theta \right)\ } \\
v_{0y}=-v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.}$.

Thus, the velocity, along $x$ and $y$, of the radar decoy is given by $\displaystyle{\left\{ \begin{array}{c}
v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ } \\
v_y\left(t\right)=-gt-v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.}$ and the position, along $x$ and $y$, of the radar decoy as it falls, is given by

\left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2-v_0{\mathrm{sin} \left(\theta \right)\ }t+h} \end{array}
\right.

The time $t_1$ spent by the decoy in the air is the time required to cover a distance $R$ at a constant horizontal velocity $v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }$. Thus, we have $x\left(t_1\right)=R$ and we solve for $t_1$ as follows

\begin{aligned}
x\left(t_1\right)=R\ \ \ \ \ &\Rightarrow \ \ \ \ \ v_0{\mathrm{cos} \left(\theta \right)\ }t_1=R \\
\\
&\Rightarrow \ \ \ \ \ t_1=\frac{R}{v_0{\mathrm{cos} \left(\theta \right)\ }}
\end{aligned}

The time of flight of the package is therefore equal to

\boxed{t_1=\frac{R}{v_0{\mathrm{cos} \left(\theta \right)\ }}=\frac{160}{100{\mathrm{cos} \left(37\right)\ }}\approx 2\ s}

The package falls a height $h$ in a time $t_1$ and therefore $t_1$ satisfies $y\left(t_1\right)=0$. This allows us to solve for $h$ as follows

\begin{aligned}
y\left(t_1\right)=0\ \ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_1-v_0{\mathrm{sin} \left(\theta \right)\ }t_1+h=0 \\
\\
&\Rightarrow \ \ \ \ \ h=\frac{1}{2}gt^2_1+v_0{\mathrm{sin} \left(\theta \right)\ }t_1
\end{aligned}

The package was therefore dropped from a height equal to

\boxed{h=\frac{1}{2}gt^2_1+v_0{\mathrm{sin} \left(\theta \right)\ }t_1=\frac{1}{2}\left(9.8\right)\cdot 2^2+100\cdot {\mathrm{sin} \left(37\right)\ }\cdot 2\approx 140\ m\approx 140\ m}

To determine the velocity vector when the package hits the ground, we must know its component at that time at time $t_1$, which are equal to

\begin{aligned}
v_x\left(t_1\right)&=v_0{\mathrm{cos} \left(\theta \right)\ }=100\cdot {\mathrm{cos} \left(37\right)\ }\approx 79.8\ m/s \\
\\
v_y\left(t_1\right)&=-gt_1-v_0{\mathrm{sin} \left(\theta \right)\ }=-9.8\cdot 2-100\cdot {\mathrm{sin} \left(37\right)\ }\approx -79.8\ m/s
\end{aligned}

Therefore, the velocity vector when the package hits the ground is

\boxed{{\overrightarrow{v}}_f=79.8\ \hat{x}\ -\ 79.8\ \hat{y}\ \ \ \ \ \left(m/s\right)}