Janet is standing at the base of a building with an outside elevator (see figure). When the elevator is at a height $h=\left(2.0\right)\left(9.8\right)\ m$, she sees a little girl throw a piece of gum from the elevator horizontally, and falling on the ground a distance $D=10\ m$ from the base of the elevator.

1. If the upward speed of the elevator is $v_E=2.0\ m/s$, what is the velocity of the gum when it leaves the girl’s hand that the girl sees? Once you specify your reference frame, you can give me either the two components of the velocity or its magnitude and direction. You can neglect the girl’s height.

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In order for Janet to see the gum leave the elevator horizontally, the little girl has to throw the gum at a downward angle so that its vertical velocity component $v_{0y}$ cancels with the upward velocity of the elevator as shown in the figure below. Thus, the initial velocity of the gum in the $y$-direction must satisfy $v_{0y}=-v_E$.

In addition, the little girl must throw the gum with a horizontal velocity component $v_{0x}$ such that it travels a distance $D$ while falling a height $h$. In the frame of reference of Janet, the gum leaves the elevator horizontally with speed $v_{0x}$ and undergoes an acceleration with components $a_x=0$ and $a_y=-g$ as it flies through the air. Its kinematic equations describing its position while it is airborne are given by

\left\{ \begin{array}{c} x\left(t\right)=v_{0x}t \\ \\ \displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+h} \end{array} \right.

The piece of gum hits the floor at a time $t_f$ such that $\displaystyle{\left\{ \begin{array}{c}
x\left(t_f\right)=D \\
y\left(t_f\right)=0 \end{array}
\right.}$ and we solve for $v_{0x}$ as follows

\begin{aligned} \left\{ \begin{array}{c} x\left(t_f\right)=D \\ \\ y\left(t_f\right)=0 \end{array} \right.\ \ \ \ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} v_{0x}t_f=D \\ \\ \displaystyle{\frac{1}{2}gt^2_f=h} \end{array} \right. \\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{v_{0x}=\frac{D}{t_f}} \\ \\ \displaystyle{t_f=\sqrt{\frac{2h}{g}}} \end{array} \right.\\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{v_{0x}=D\sqrt{\frac{g}{2h}}} \\ \\ \displaystyle{t_f=\sqrt{\frac{2h}{g}}} \end{array} \right. \end{aligned}

Thus, the horizontal component of the initial velocity of the gum must be equal to

v_{0x}=D\sqrt{\frac{g}{2h}}=10\sqrt{\frac{9.8}{2\cdot 2\cdot 9.8}}=5\ m/s

In the frame of reference of the little girl, the piece of gum must be thrown with an initial velocity equal to

\boxed{{\overrightarrow{v}}_0=v_{0x}\ \ \hat{x}-v_E\ \ \hat{y}=5\ \ \hat{x}-2\ \ \hat{y}\ \ \ \left(m/s\right)}

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