Question

(u5) suppose brett throws a baseball into the air, which follows a parabolic path and where h (height) is in feet and t (time) is in seconds. brett releases the ball from a height of 6 feet above the ground. at one second the ball’s height is 69 feet, at 2 seconds the ball’s height is 100 feet, at 4 seconds the ball’s height is 66, and at 5 seconds it is 1 foot off the ground. what points would you use to find the quadratic equation? what is the quadratic equation? at the same time brett releases the ball and jeremy shoots an arrow from the top of the baseball stands. the path of the arrow can be represented by h(t) = (23/2)t + 20. what is last intersection of the arrow and baseball.

Explanation:

Step1: Identify the points

We know that for a quadratic equation representing height \( h \) as a function of time \( t \), the points are in the form \( (t, h) \). From the problem:

• When \( t = 0 \) (release time), \( h = 6 \), so point \( (0, 6) \)
• When \( t = 1 \), \( h = 69 \), so point \( (1, 69) \)
• When \( t = 2 \), \( h = 100 \), so point \( (2, 100) \)
• When \( t = 4 \), \( h = 66 \), so point \( (4, 66) \)
• When \( t = 5 \), \( h = 1 \), so point \( (5, 1) \)

Step2: Find the quadratic equation

A quadratic equation is of the form \( h(t)=at^{2}+bt + c \). We can use the point \( (0, 6) \) to find \( c \) first. Substituting \( t = 0 \) and \( h = 6 \) into \( h(t)=at^{2}+bt + c \), we get \( 6=a(0)^{2}+b(0)+c\), so \( c = 6 \).

Now we have \( h(t)=at^{2}+bt + 6 \). Substitute \( (1, 69) \): \( 69=a(1)^{2}+b(1)+6\Rightarrow a + b=63\) (Equation 1)

Substitute \( (2, 100) \): \( 100=a(2)^{2}+b(2)+6\Rightarrow 4a + 2b=94\Rightarrow 2a + b = 47\) (Equation 2)

Subtract Equation 1 from Equation 2: \( (2a + b)-(a + b)=47 - 63\Rightarrow a=-16\)

Substitute \( a=-16 \) into Equation 1: \( - 16+b=63\Rightarrow b = 79\)

So the quadratic equation is \( h(t)=-16t^{2}+79t + 6 \)

Step3: Find the intersection of the two functions

We need to solve \( -16t^{2}+79t + 6=\frac{23}{2}t+20 \)

Multiply through by 2 to eliminate the fraction: \( - 32t^{2}+158t + 12 = 23t+40 \)

Rearrange to standard quadratic form: \( -32t^{2}+135t - 28 = 0\) or \( 32t^{2}-135t + 28 = 0 \)

Using the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) where \( a = 32 \), \( b=-135 \), \( c = 28 \)

\( t=\frac{135\pm\sqrt{(-135)^{2}-4\times32\times28}}{2\times32}=\frac{135\pm\sqrt{18225 - 3584}}{64}=\frac{135\pm\sqrt{14641}}{64}=\frac{135\pm121}{64} \)

We get two solutions: \( t_1=\frac{135 + 121}{64}=\frac{256}{64}=4 \) and \( t_2=\frac{135-121}{64}=\frac{14}{64}=\frac{7}{32} \)

The last intersection (larger \( t \)) is at \( t = 4 \)

Answer:

• Points: \((0, 6)\), \((1, 69)\), \((2, 100)\), \((4, 66)\), \((5, 1)\)
• Quadratic equation: \( h(t)=-16t^{2}+79t + 6 \)
• Last intersection: \( t = 4 \) (and the corresponding height can be found by substituting \( t = 4 \) into either function, e.g., \( h(4)=\frac{23}{2}\times4+20=46 + 20 = 66 \), so the point is \((4, 66)\))