# Question #6903b

May 27, 2017

The height of the cliff is $= 15 m$, the maximum height is $= 16 m$
The speed is $= 2 \left(1 - t\right)$, After $4 s$, $v = - 6 m {s}^{-} 1$,The speed at $10 m$ is $= - 2 \sqrt{6}$

#### Explanation:

The equation is

$h \left(t\right) = 15 + 2 t - {t}^{2}$

The height of the cliff is when $t = 0$

$h \left(0\right) = 15 + 0 - 0 = 15 m$

To calculate the maximum height, we calculate the derivative

$h ' \left(t\right) = 2 - 2 t$

$h ' \left(t\right) = 0$ when $2 - 2 t = 0$, $\implies$, $t = 1$

Therefore,

${h}_{\max} = 15 + 2 - 1 = 16 m$

The speed is the derivative of the position

$v \left(t\right) = h ' \left(t\right) = 2 - 2 t$

When $t = 4$

$v \left(4\right) = 2 - 8 = - 6 m {s}^{-} 1$, the minus sign indicates that the ball is going down.

When $h = 10 m$

$10 = 15 + 2 t - {t}^{2}$

${t}^{2} - 2 t - 5 = 0$

$t = \frac{2 \pm \sqrt{4 - \left(4\right) \left(1\right) \left(- 5\right)}}{2} = \frac{2 \pm \sqrt{24}}{2} = \frac{2 \pm 2 \sqrt{6}}{2} = 1 \pm \sqrt{6}$

We neglect the negative solution

$t = 1 + \sqrt{6}$

Therefore,

The speed at $t = 1 + \sqrt{6}$is

$v \left(1 + \sqrt{6}\right) = 2 \left(1 - \left(1 + \sqrt{6}\right)\right) = - 2 \sqrt{6}$
graph{15+2x-x^2 [-0.17, 35.87, -0.72, 17.3]}