Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. a baseball is thrown upward at a velocity of 36 feet per second from a height of 7 feet. the equation to represent the height (h) after t seconds is h = -16t² + 36t + 7. how long before the ball hits the ground? (round to the nearest hundredth of a second.) the solution is

Explanation:

Step1: Set \( h = 0 \)

We need to find when the ball hits the ground, so set \( h = 0 \) in the equation \( h=-16t^{2}+36t + 7 \). We get the quadratic equation \( -16t^{2}+36t + 7=0 \). Multiply both sides by - 1 to make it easier: \( 16t^{2}-36t - 7 = 0 \)

Step2: Use quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \) (here \( x=t \), \( a = 16 \), \( b=-36 \), \( c=-7 \)), the quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)

First, calculate the discriminant \( D=b^{2}-4ac \)
\( D=(-36)^{2}-4\times16\times(-7) \)
\( D = 1296+448 \)
\( D=1744 \)

Then, find \( t \)
\( t=\frac{36\pm\sqrt{1744}}{32} \)

We have two solutions:
\( t_1=\frac{36+\sqrt{1744}}{32}\approx\frac{36 + 41.76}{32}=\frac{77.76}{32}\approx2.43 \)
\( t_2=\frac{36-\sqrt{1744}}{32}\approx\frac{36 - 41.76}{32}=\frac{- 5.76}{32}\approx - 0.18 \)

Since time cannot be negative, we discard \( t_2 \)

Answer:

\( 2.43 \)