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]}