Question

video lesson 4.2 example 6 - real-life application.

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) ).

Explanation:

Step1: For part b

We know that when \(t = 1\), \(h=46\) and when \(t = 0\), \(h = 2\). The height \(h = 40\) is between \(t=0\) and \(t = 1\).
Let's assume \(h=-16t^{2}+60t + 2=40\), then \(-16t^{2}+60t+2 - 40=0\), or \(-16t^{2}+60t - 38 = 0\), multiply by \(- 1\) gives \(16t^{2}-60t + 38=0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 16\), \(b=-60\), \(c = 38\). But using the table - based estimate:
We can also use a linear - approximation. The change in height from \(t = 0\) (\(h(0)=2\)) to \(t = 1\) (\(h(1)=46\)) is \(\Delta h=46 - 2=44\) in \(\Delta t = 1\) second. We want to find \(t\) when \(h = 40\). The change from \(h = 2\) to \(h = 40\) is \(\Delta h'=40 - 2=38\). Let \(t=x\), then \(\frac{38}{44}\approx x\), \(x\approx0.86\)

Step2: For part c

We know that when \(t = 3\), \(h = 38\) and when \(t = 4\), \(h=-14\). The height \(h = 0\) is between \(t = 3\) and \(t = 4\).
Let's assume \(h=-16t^{2}+60t + 2 = 0\). Using the quadratic formula \(t=\frac{-60\pm\sqrt{60^{2}-4\times(-16)\times2}}{2\times(-16)}=\frac{-60\pm\sqrt{3600 + 128}}{-32}=\frac{-60\pm\sqrt{3728}}{-32}=\frac{-60\pm61.06}{-32}\). We take the positive root \(t=\frac{-60 + 61.06}{-32}\) (wrong, correct formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=-16\), \(b = 60\), \(c = 2\)) \(t=\frac{-60\pm\sqrt{60^{2}-4\times(-16)\times2}}{2\times(-16)}=\frac{-60\pm\sqrt{3600 + 128}}{-32}=\frac{-60\pm\sqrt{3728}}{-32}=\frac{-60\pm61.06}{-32}\). The positive root \(t=\frac{-60 - 61.06}{-32}\approx3.78\). Using the table - based estimate:
The change in height from \(t = 3\) (\(h(3)=38\)) to \(t = 4\) (\(h(4)=-14\)) is \(\Delta h=-14 - 38=-52\) in \(\Delta t = 1\) second. We want to find \(t\) when \(h = 0\). The change from \(h = 38\) to \(h = 0\) is \(\Delta h'=-38\). Let \(t = 3 + y\), then \(\frac{-38}{-52}\approx y\), \(y\approx0.73\), so \(t\approx3.73\)

Answer:

b. The height of the soccer - ball is approximately \(40\) feet at \(t\approx0.86\) seconds.
c. The soccer - ball hits the ground at \(t\approx3.7\) seconds.