Daniel kicks a soccer ball up into the air with an initial upward velocity of 63 feet per second. The ball is 2 feet above the ground when it is kicked. Write an equation to model the situation. What is the maximum height of the ball? How long will it take the ball to reach its maximum height? What is the range of the function?

Guest May 4, 2015

#2**+10 **

**Daniel kicks a soccer ball up into the air with an initial upward velocity of 63 feet per second. The ball is 2 feet above the ground when it is kicked. Write an equation to model the situation.**

**I. Write an equation to model the situation.**

$$\small{\text{$~\boxed{~h(t) = \dfrac{g}{2}\cdot t^2 + v\cdot t + h_0 \qquad g = -32 ~\frac{ft.}{sec^2} \qquad v = 63~\frac{ft.}{sec} \qquad h_0 = 2~ft. ~}

~$}}$$

The height of the soccer ball at any given time, t, can be calculated according to the function,

$$\small\text{$~h(t) = -16\cdot t^2 + 63\cdot t + 2 ~$}}$$ , where time is measured in seconds and the height is measured in feet.

This is a parabola $$\small{\text{$~y = a\cdot x^2+b\cdot x+c~$}}$$.

**II. How long will it take the ball to reach its maximum height ?**

The vertex of the parabola is $$\small{\text{$~x=t=\frac{-b}{2a} = \frac{-63}{2\cdot(-16)}=1.96875 ~\rm{sec.} ~$}}$$

**III. What is the maximum height of the ball ?**

**$$\small\text{$~ \begin{array}{rcl} h(1.96875~\rm{sec.}) &=& -16\cdot {1.96875}^2 + 63\cdot 1.96875 + 2\\ &=& -62.015625~\rm{ft.} +124.03125~\rm{ft.} +2 ~\rm{ft.}\\ h(1.96875~\rm{sec.}) &=& 64.015625~\rm{ft.} \end{array} ~$}}$$**

heureka
May 4, 2015

#1**+10 **

Since the only force acting on the ball after it has been kicked is gravity we can use the constant acceleration equations:

v^{2} = u^{2} - 2*g*s ...(1) where v = final velocity, u = initial velocity, g = 32 ft/sec^{2} and s = distance travelled

and

v = u - g*t ...(2) where t = time

Maximum height is when v = 0, so, using (1):

0 = 63^{2} - 2*32*s

s = 63^{2}/(2*32) ≈ 62 ft

But the ball was kicked from astarting height of 2 ft, so maximum height is 64 ft.

Using (2) we have

0 = 63 - 32*t

t = 63/32 ≈ 1.97 seconds

Alan
May 4, 2015

#2**+10 **

Best Answer

**Daniel kicks a soccer ball up into the air with an initial upward velocity of 63 feet per second. The ball is 2 feet above the ground when it is kicked. Write an equation to model the situation.**

**I. Write an equation to model the situation.**

$$\small{\text{$~\boxed{~h(t) = \dfrac{g}{2}\cdot t^2 + v\cdot t + h_0 \qquad g = -32 ~\frac{ft.}{sec^2} \qquad v = 63~\frac{ft.}{sec} \qquad h_0 = 2~ft. ~}

~$}}$$

The height of the soccer ball at any given time, t, can be calculated according to the function,

$$\small\text{$~h(t) = -16\cdot t^2 + 63\cdot t + 2 ~$}}$$ , where time is measured in seconds and the height is measured in feet.

This is a parabola $$\small{\text{$~y = a\cdot x^2+b\cdot x+c~$}}$$.

**II. How long will it take the ball to reach its maximum height ?**

The vertex of the parabola is $$\small{\text{$~x=t=\frac{-b}{2a} = \frac{-63}{2\cdot(-16)}=1.96875 ~\rm{sec.} ~$}}$$

**III. What is the maximum height of the ball ?**

**$$\small\text{$~ \begin{array}{rcl} h(1.96875~\rm{sec.}) &=& -16\cdot {1.96875}^2 + 63\cdot 1.96875 + 2\\ &=& -62.015625~\rm{ft.} +124.03125~\rm{ft.} +2 ~\rm{ft.}\\ h(1.96875~\rm{sec.}) &=& 64.015625~\rm{ft.} \end{array} ~$}}$$**

heureka
May 4, 2015