Question

A girl holds an arrow in her hands, ready to shoot the arrow at a dragon who is just outside her arrow’s range of 45 meters. The girl then slides down a cliffside, giving her a velocity of 8.5 m/s at an angle of 45 below the horizontal. She points her bow 30 degrees above the horizontal and shoots, the arrow leaves the bow with a velocity of 45 m/s. The girl is 55 meters away from the dragon horizontally. How far above the girl is the dragon? (Assuming the arrow hits.)I already figured out how long it takes for the arrow to reach the dragon (4.72 seconds) but I don’t know how to calculate the distance between the girl and the dragon. Please help!

Answer 1

First, assume that the arrow leaves the bow with a velocity of 45 m/s above the horizontal with respect to the bow.

Since the bow is moving at 8.5 m/s 45º below the horizontal, find the initial velocity of the arrow with respect to the ground:

$\vec{v}_{AG}=\vec{v}_{AB}+\vec{v}_{BG}$

This equation reads:

The velocity of the arrow with respect to the ground is equal to the velocity of the arrow with respect to the bow plus the velocity of the bow with respect to the gound.

Notice that this is a vector equation. Then, the vertical and horizontal components of the velocities must be added separately:

$\begin{gathered} v_{AG-x}=v_{AB-x}+v_{BG-x} \\ v_{AG-y}=v_{AB-y}+v_{BG-y} \end{gathered}$

Find the vertical and horizontal components of the velocity of the arrow with respect to the bow and the velocity of the bow with respect to the ground:

$\begin{gathered} v_{AB-x}=v_{AB}\cos (\theta) \\ =45\frac{m}{s}\cdot\cos (30º) \\ =38.97\frac{m}{s} \end{gathered}$

$\begin{gathered} v_{AB-y}=v_{AB}\sin (\theta) \\ =45\frac{m}{s}\sin (30º) \\ =22.5\frac{m}{s} \end{gathered}$

Similarly, for the velocity of the bow with respect to the ground:

$\begin{gathered} v_{BG-x}=6.01\frac{m}{s} \\ v_{BG-y}=-6.01\frac{m}{s} \end{gathered}$

Then, the vertical and horizontal components of the initial velocity of the arrow with respect to the ground, are:

$\begin{gathered} v_{AG-x}=38.97\frac{m}{s}+6.01\frac{m}{s}=44.98\frac{m}{s} \\ \\ v_{AG-y}=22.5\frac{m}{s}-6.01\frac{m}{s}=16.49\frac{m}{s} \end{gathered}$

Use the horizontal component of the velocity to find how long it takes for the arrow to travel a horizontal distance x of 55 meters. Then, use that time to find the vertical position of the arrow.

Since the horizontal movement of the arrow is uniform, then:

$v_{AG-x}=\frac{x}{t}_{}$

Isolate t and substitute x=55m, v_{AG-x}=44.98 m/s:

$\begin{gathered} t=\frac{x}{v_{AG-x}} \\ =\frac{55m}{44.98\frac{m}{s}} \\ =1.2227s \end{gathered}$

The vertical motion of the arrow is a uniformly accelerated motion. Then, the vertical position is given by:

$y=v_{AG-y}t-\frac{1}{2}gt^2$

Replace v_{AG-y}=16.49 m/s, t=1.2227s and g=9.81 m/s^2 to find the vertical position of the arrow when the horizontal position is 55 meters. This matches the elevation of the dragon with respect to the girl when the girl shoots:

$\begin{gathered} y=(16.49\frac{m}{s})(1.2227s)-\frac{1}{2}(9.81\frac{m}{s^2})(1.2227s)^2 \\ =12.829\ldots m \\ \approx12.8m \end{gathered}$

Therefore, the dragon is 12.8 meters above the girl when the arrow is shoot.