Question #6903b

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Answer 1 · 2017-05-27T06:25:46

The height of the cliff is $=15m$ , the maximum height is $=16m$
The speed is $=2(1-t)$ , After $4s$ , $v=-6ms^-1$ ,The speed at $10m$ is $=-2sqrt6$

The equation is

$h(t)=15+2t-t^2$

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

$h(0)=15+0-0=15m$

To calculate the maximum height, we calculate the derivative

$h'(t)=2-2t$

$h'(t)=0$ when $2-2t=0$ , $=>$ , $t=1$

Therefore,

$h_(max)=15+2-1=16m$

The speed is the derivative of the position

$v(t)=h'(t)=2-2t$

When $t=4$

$v(4)=2-8=-6ms^-1$ , the minus sign indicates that the ball is going down.

When $h=10m$

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

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

We solve this quadratic equation,

$t=(2+-sqrt(4-(4)(1)(-5)))/(2)=(2+-sqrt24)/2=(2+-2sqrt6)/2=1+-sqrt6$

We neglect the negative solution

$t=1+sqrt6$

Therefore,

The speed at $t=1+sqrt6$ is

$v(1+sqrt6)=2(1-(1+sqrt6))=-2sqrt6$
graph{15+2x-x^2 [-0.17, 35.87, -0.72, 17.3]}