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 i

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 ?

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

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

v² = u² - 2*g*s   ...(1)    where v = final velocity, u = initial velocity, g = 32 ft/sec² and s = distance travelled

and

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

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

0 = 63² - 2*32*s

s = 63²/(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

Thanks Alan and Heureka,

Anon, are you happy with these 2 great answers or were you needing a calculus answer?