Question

technology required. two rocks are launched straight up in the air. the height of rock a is given by the function ( f ), where ( f(t) = 4 + 30t - 16t^2 ). the height of rock b is given by ( g ), where ( g(t) = 5 + 20t - 16t^2 ). in both functions, ( t ) is time measured in seconds and height is measured in feet. use graphing technology to graph both equations. determine which rock hits the ground first and explain how you know.

Explanation:

Step1: Find when Rock A hits ground

Set \( f(t) = 0 \), so \( 4 + 30t - 16t^2 = 0 \). Using quadratic formula \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) for \( ax^2+bx+c = 0 \), here \( a=-16 \), \( b = 30 \), \( c = 4 \).
\( t=\frac{-30\pm\sqrt{30^2 - 4(-16)(4)}}{2(-16)}=\frac{-30\pm\sqrt{900 + 256}}{-32}=\frac{-30\pm\sqrt{1156}}{-32}=\frac{-30\pm34}{-32} \).
Positive solution: \( t=\frac{-30 + 34}{-32}=\frac{4}{-32}=-0.125 \) (discard) or \( t=\frac{-30 - 34}{-32}=\frac{-64}{-32}=2 \) seconds.

Step2: Find when Rock B hits ground

Set \( g(t) = 0 \), so \( 5 + 20t - 16t^2 = 0 \). Using quadratic formula, \( a=-16 \), \( b = 20 \), \( c = 5 \).
\( t=\frac{-20\pm\sqrt{20^2 - 4(-16)(5)}}{2(-16)}=\frac{-20\pm\sqrt{400 + 320}}{-32}=\frac{-20\pm\sqrt{720}}{-32}=\frac{-20\pm12\sqrt{5}}{-32} \).
Positive solution: \( t=\frac{-20 - 12\sqrt{5}}{-32}\approx\frac{-20 - 26.83}{-32}\approx\frac{-46.83}{-32}\approx1.463 \) seconds (other solution negative, discard).

Step3: Compare times

Rock A hits ground at \( t = 2 \) s, Rock B at \( t\approx1.463 \) s. Since \( 1.463 < 2 \), Rock B hits first.

Answer:

Rock B hits the ground first. Rock A hits at \( t = 2 \) seconds and Rock B hits at approximately \( 1.463 \) seconds, so Rock B's time to ground is less.