Question

1. a soccer player kicks a soccer ball 2 feet above the ground with an initial vertical velocity of 60 feet per second. the function ( h = - 16 t ^ { 2 } + 60 t + 2 ) represents the height ( h ) (in feet) of the soccer ball after ( t ) seconds.

a. find the height of the soccer ball each second after it is kicked.

seconds, ( t )	0	1	2	3	4	5

b. use the results in part (a) to estimate when the height of the soccer ball is 40 feet.

c. use the results in part (a) to estimate when the soccer ball hits the ground ( ( h = 0 ) ).

d. use a graphing calculator to check your results from parts (b) and (c).

Explanation:

Step1: Analyze part (b)

We know the function \(h = - 16t^{2}+60t + 2\). The height values in part (a) are: at \(t = 0\), \(h=2\); \(t = 1\), \(h = 46\); \(t=2\), \(h = 58\); \(t = 3\), \(h=38\). Since the height changes from \(46\) (at \(t = 1\)) to \(38\) (at \(t = 3\)), and we want to find when \(h = 40\). We can assume the height is a quadratic function (parabola). The height is decreasing from \(t=2\) to \(t = 3\). Let's use the linear - approximation between \(t = 1\) (\(h = 46\)) and \(t=3\) (\(h = 38\)). The slope of the line between the points \((1,46)\) and \((3,38)\) is \(m=\frac{38 - 46}{3 - 1}=\frac{-8}{2}=-4\). The equation of the line using the point - slope form \(y - y_1=m(x - x_1)\) (taking \((x_1,y_1)=(1,46)\)) is \(y-46=-4(x - 1)\), or \(y=-4x + 50\). Set \(y = 40\), then \(40=-4x + 50\), \(4x=10\), \(x = 2.5\). Also, since the function is symmetric about \(t=-\frac{b}{2a}=-\frac{60}{2\times(-16)}=\frac{60}{32}=\frac{15}{8}=1.875\). The height at \(t = 0\) is \(h = 2\), and the function is increasing from \(t = 0\) to \(t=1.875\). We can check the value of the function at \(t=0.1\): \(h=-16\times(0.1)^{2}+60\times(0.1)+2=-16\times0.01 + 6+2=7.84\); \(t = 0.5\): \(h=-16\times(0.5)^{2}+60\times(0.5)+2=-16\times0.25+30 + 2=28\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(h=-16t^{2}+60t + 2\) and \(h = 40\), we have \(-16t^{2}+60t + 2=40\), or \(16t^{2}-60t + 38 = 0\), \(8t^{2}-30t + 19=0\). \(t=\frac{30\pm\sqrt{900 - 608}}{16}=\frac{30\pm\sqrt{292}}{16}=\frac{30\pm2\sqrt{73}}{16}\approx\frac{30\pm16.248}{16}\). \(t_1=\frac{30 + 16.248}{16}\approx2.89\), \(t_2=\frac{30 - 16.248}{16}\approx0.86\).

Step2: Analyze part (c)

We want to find \(t\) when \(h=-16t^{2}+60t + 2 = 0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a=-16\), \(b = 60\), \(c = 2\). \(t=\frac{-60\pm\sqrt{60^{2}-4\times(-16)\times2}}{2\times(-16)}=\frac{-60\pm\sqrt{3600 + 128}}{-32}=\frac{-60\pm\sqrt{3728}}{-32}=\frac{-60\pm61.06}{-32}\). We take the positive root \(t=\frac{-60 + 61.06}{-32}\) (rejected) or \(t=\frac{-60-61.06}{-32}\approx\frac{-121.06}{-32}\approx3.78\). Looking at the table: at \(t = 3\), \(h = 38\); at \(t=4\), \(h=-14\). The function crosses the \(t\) - axis (where \(h = 0\)) between \(t = 3\) and \(t = 4\).

Answer:

b. The height of the soccer ball is \(40\) feet at approximately \(t = 0.86\) seconds and \(t = 2.89\) seconds.
c. The soccer ball hits the ground at approximately \(t = 3.78\) seconds.