Question

imagine that the function $h(t) = -0.16t^2 + 1.7t + 18$ models the height (in feet) of an \angry bird\ shot into the sky as a function of time (seconds). use this function and a scientific calculator to answer the following questions. round all answers to two decimal places
a) how high above the ground was the bird when it was launched?
feet
b) after how many seconds until it reach its highest point?
seconds
c) how high is the highest point?
feet
d) after how many seconds does the angry bird hit the ground?
seconds

Explanation:

Part (a)

Step 1: Understand the launch time

When the bird is launched, the time \( t = 0 \) seconds. We need to find the height \( h(0) \) by substituting \( t = 0 \) into the function \( h(t)=- 0.16t^{2}+1.7t + 18 \).

Step 2: Substitute \( t = 0 \)

\( h(0)=-0.16(0)^{2}+1.7(0)+18 \)
\( h(0)=0 + 0+18=18 \)

Part (b)

Step 1: Recall the formula for the vertex of a parabola

For a quadratic function in the form \( y = ax^{2}+bx + c \), the \( x \)-coordinate of the vertex (which gives the time at the highest point for \( h(t) \)) is given by \( t=-\frac{b}{2a} \)

Step 2: Identify \( a \) and \( b \)

In the function \( h(t)=-0.16t^{2}+1.7t + 18 \), \( a=- 0.16 \) and \( b = 1.7 \)

Step 3: Calculate \( t \)

\( t=-\frac{1.7}{2\times(-0.16)}=-\frac{1.7}{- 0.32}=\frac{1.7}{0.32}\approx5.31 \)

Part (c)

Step 1: Use the time from part (b)

We know that the time at the highest point is \( t\approx5.31 \) seconds. We substitute this value into the function \( h(t) \) to find the maximum height.

Step 2: Substitute \( t = 5.31 \) into \( h(t) \)

\( h(5.31)=-0.16(5.31)^{2}+1.7(5.31)+18 \)
First, calculate \( (5.31)^{2}=28.1961 \)
Then, \( - 0.16\times28.1961=-4.511376 \)
\( 1.7\times5.31 = 9.027 \)
Now, \( h(5.31)=-4.511376 + 9.027+18=22.515624\approx22.52 \)

Part (d)

Step 1: Set \( h(t) = 0 \)

We need to solve the quadratic equation \( -0.16t^{2}+1.7t + 18=0 \) for \( t \). The quadratic formula for a quadratic equation \( ax^{2}+bx + c = 0 \) is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)

Step 2: Identify \( a \), \( b \) and \( c \)

Here, \( a=-0.16 \), \( b = 1.7 \) and \( c = 18 \)

Step 3: Calculate the discriminant \( D=b^{2}-4ac \)

\( D=(1.7)^{2}-4\times(-0.16)\times18 \)
\( D = 2.89+11.52=14.41 \)

Step 4: Calculate \( t \) using the quadratic formula

\( t=\frac{-1.7\pm\sqrt{14.41}}{2\times(-0.16)}=\frac{-1.7\pm3.8}{-0.32} \)

We have two solutions:

• \( t_1=\frac{-1.7 + 3.8}{-0.32}=\frac{2.1}{-0.32}=-6.5625 \) (we discard this negative solution since time cannot be negative)
• \( t_2=\frac{-1.7-3.8}{-0.32}=\frac{-5.5}{-0.32}=17.1875\approx17.19 \)

Answer:

s:
a) \(\boldsymbol{18.00}\) feet
b) \(\boldsymbol{5.31}\) seconds
c) \(\boldsymbol{22.52}\) feet
d) \(\boldsymbol{17.19}\) seconds