Question

a person shoots an arrow vertically into the air from a height of 6 feet with an initial velocity of 96 feet per second. the height, h, in feet above the ground, at any time, t (in seconds), is modeled by h(t) = 6 + 96t - 16t². a) determine the maximum height the arrow will attain. 150 feet b) approximately when will the arrow reach the ground? after approximately seconds the arrow will reach the ground (type an integer or decimal rounded to the nearest hundredth as needed.)

Explanation:

Step1: Set the height function to zero

We need to find when \( h(t) = 0 \), so we set up the equation \( 0 = 6 + 96t - 16t^{2} \). This is a quadratic equation in the form \( at^{2}+bt + c = 0 \), where \( a=- 16 \), \( b = 96 \), and \( c = 6 \).

Step2: Use the quadratic formula

The quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Substituting the values of \( a \), \( b \), and \( c \) into the formula, we get:

$$ t=\frac{-96\pm\sqrt{96^{2}-4\times(-16)\times6}}{2\times(-16)} $$

First, calculate the discriminant \( D=b^{2}-4ac=96^{2}-4\times(-16)\times6=9216 + 384=9600 \)

Then,

$$ t=\frac{-96\pm\sqrt{9600}}{-32}=\frac{-96\pm40\sqrt{6}}{-32}=\frac{96\mp40\sqrt{6}}{32}=\frac{12\mp5\sqrt{6}}{4} $$

We have two solutions for \( t \):
\( t_{1}=\frac{12 + 5\sqrt{6}}{4}\approx\frac{12+5\times2.45}{4}=\frac{12 + 12.25}{4}=\frac{24.25}{4} = 6.0625\approx6.06 \)
\( t_{2}=\frac{12-5\sqrt{6}}{4}\approx\frac{12 - 12.25}{4}=\frac{- 0.25}{4}=- 0.0625 \)

Since time cannot be negative, we discard \( t_{2} \). So the arrow reaches the ground after approximately \( 6.06 \) seconds.

Answer:

6.06