Certainly! Let's analyze the system of equations provided:
[tex]\[
\left\{\begin{array}{l}
h = 8 + 2t \\
h = 4 + 32t - 16t^2
\end{array}\right.
\][/tex]
The first equation, [tex]\( h = 8 + 2t \)[/tex], describes the height [tex]\( h \)[/tex] of the target as it is lifted into the air by a cable at a constant speed. The term [tex]\( 2t \)[/tex] indicates that the target is rising at a constant speed of 2 feet per second, starting from an initial height of 8 feet.
The second equation, [tex]\( h = 4 + 32t - 16t^2 \)[/tex], describes the height [tex]\( h \)[/tex] of the arrow after it has been fired upwards. This is a quadratic equation representing the motion of the arrow under the influence of gravity. In this equation:
- The term [tex]\( 4 \)[/tex] represents the initial height from which the arrow was launched.
- The term [tex]\( 32t \)[/tex] represents the initial upward velocity component.
- The term [tex]\( -16t^2 \)[/tex] accounts for the downward acceleration due to gravity (with the factor [tex]\( -16 \)[/tex] derived from [tex]\(-\frac{1}{2} \cdot 32 \)[/tex], where gravity [tex]\( g \)[/tex] is [tex]\( 32 \)[/tex] ft/s²).
To determine the initial upward velocity [tex]\( v_0 \)[/tex] of the arrow, we compare the general equation of motion for the arrow's height with the given equation:
[tex]\[ h = h_0 + v_0 t - \frac{1}{2} g t^2 \][/tex]
Matching this with [tex]\( h = 4 + 32t - 16t^2 \)[/tex], we identify:
- [tex]\( h_0 = 4 \)[/tex]
- [tex]\( v_0 = 32 \)[/tex] ft/s
- The coefficient of [tex]\( t^2 \)[/tex] matches the term [tex]\( -\frac{1}{2} g \cdot t^2 \)[/tex], with [tex]\( g = 32 \)[/tex] ft/s².
Thus, the initial upward velocity [tex]\( v_0 \)[/tex] of the arrow is [tex]\( 32 \)[/tex] ft/s.
So, the correct statement describing the scenario is:
The arrow is fired with an initial upward velocity of [tex]\( 32 \)[/tex] ft/s.
