The height in feet of a golf ball hit into the air is given by h= -16t^2 + 64t, where t is the number of seconds elapsed since the ball was hit. What is the maximum height of the ball?

Mar 20, 2017

$64$ feet

Explanation:

$h = - 16 {t}^{2} + 64 t$

maximum or minimum height when $h ' = 0$

$h ' = - 32 t + 64$

when $h ' = 0 , - 32 t + 64 = 0$

$32 t = 64$, $t = \frac{64}{32} = 2$

$h ' ' = - 32$, and since $h ' '$ has a -ve value, therefore $h$ is maximum when $t = 2$

when t =2,
$h = - 16 {\left(2\right)}^{2} + 64 \left(2\right)$
$h = - 16 {\left(2\right)}^{2} + 64 \left(2\right)$
$h = - 16 \left(4\right) + 64 \left(2\right) = - 64 + 128 = 64$

Mar 20, 2017

$64 \text{ feet}$

Explanation:

Find the roots of the parabola by setting h = 0

$\Rightarrow - 16 {t}^{2} + 64 t = 0$

Factorising gives.

$- 16 t \left(t - 4\right) = 0$

$\Rightarrow t = 0 \text{ or } t = 4$

Since the parabola is symmetrical the axis of symmetry will pass through the maximum. The axis of symmetry is positioned at the mid- point of the roots.

$\Rightarrow \textcolor{red}{t = 2}$

$\Rightarrow \text{max. height } = - 16 {\textcolor{red}{\left(2\right)}}^{2} + 64 \left(\textcolor{red}{2}\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = \left(- 16 \times 4\right) + \left(64 \times 2\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = - 64 + 128$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = 64$

$\Rightarrow \text{maximum height of ball " =64" feet}$