Question

as part of a lesson on gravity, ms. keller tossed an orange into the air from the roof of the school. the orange had an initial velocity of 3 meters per second and an initial height of 9 meters. the equation $h(t) = -4.9t^{2} + 3t + 9$ models the height of the orange in meters after t seconds. about how many seconds does it take for the orange to hit the ground? round your answer to the nearest tenth of a second. seconds submit

Explanation:

Step1: Set height to 0

We need to find when the orange hits the ground, so set \( h(t) = 0 \). The equation becomes \( 0=-4.9t^{2}+3t + 9 \). This is a quadratic equation in the form \( at^{2}+bt + c = 0 \), where \( a=-4.9 \), \( b = 3 \), and \( c = 9 \).

Step2: Use quadratic formula

The quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Substitute the values of \( a \), \( b \), and \( c \) into the formula:
First, calculate the discriminant \( D=b^{2}-4ac=(3)^{2}-4\times(-4.9)\times9 = 9 + 176.4=185.4 \)
Then, \( t=\frac{-3\pm\sqrt{185.4}}{2\times(-4.9)}=\frac{-3\pm13.616}{-9.8} \)
We have two solutions:
\( t_1=\frac{-3 + 13.616}{-9.8}=\frac{10.616}{-9.8}\approx - 1.08 \) (we discard this negative solution because time cannot be negative)
\( t_2=\frac{-3-13.616}{-9.8}=\frac{-16.616}{-9.8}\approx1.7 \)

Answer:

\( 1.7 \)