Question

a grasshopper jumps off of a tree stump. the height, in feet, of the grasshopper above the ground after t seconds is modeled by the function
$h(t) = -t^2 + \frac{4}{3}t + \frac{1}{4}$
after how many seconds will the grasshopper land on the ground?

Explanation:

Step1: Set height to zero

To find when the grasshopper lands, set \( h(t) = 0 \):
\( -t^2 + \frac{4}{3}t + \frac{1}{4} = 0 \)

Step2: Multiply by -12 to eliminate fractions

Multiply equation by \(-12\) (LCM of 3,4) to simplify:
\( 12t^2 - 16t - 3 = 0 \)

Step3: Apply quadratic formula

For \( ax^2 + bx + c = 0 \), \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 12 \), \( b = -16 \), \( c = -3 \):
Discriminant: \( (-16)^2 - 4(12)(-3) = 256 + 144 = 400 \)
\( t = \frac{16 \pm \sqrt{400}}{24} = \frac{16 \pm 20}{24} \)

Step4: Solve for t (positive solution)

Two solutions:
\( t = \frac{16 + 20}{24} = \frac{36}{24} = \frac{3}{2} \) (positive, valid)
\( t = \frac{16 - 20}{24} = \frac{-4}{24} = -\frac{1}{6} \) (discarded, time can’t be negative)

Answer:

\(\frac{3}{2}\)