How do I solve this?

### 1 Answer | Add Yours

We have given

`h=-15.9t^2+110.48t+3.02`

`` At maximum height , its velocity should be zero.

So differenciate with respect to t

`(dh)/(dt)=-2xx15.9t+110.48`

`(dh)/(dt)=0`

`if`

`-31.8t+110.48=0`

`t=110.48/31.8`

`t=3.47 sec.`

`(d^2h)/(dt^2)=-31.8<0`

Thus at t=3.47 sec , we have maximum height.

maximum height= -15.9 (3.47)^2+110.48(3.47)+3.02

=194.47 feet.

The distance ball will have to cover after attaining maximum height ,before it catched= (194.47-6)

=188.47 feet.

Substitute h=188.47 feet in given equation

h=-15.9t^2+110.48t+3.02

Thus we have

`188.47=-15.9t^2+110.48t+3.02`

`15.9t^2-110.48t+185.45=0`

solving the quadratic equation

`t=(110.48+-sqrt(110.48^2-4xx185.45xx15.9))/(2xx15.9)`

`t=(110.48+-20.28)/31.8`

`t=4.11 or 2.84`

To attained maximum height ball took 3.47 sec. so in returnjourney it should take less time tahn 3.47.

Thus t=2.84 sec.

Thus outfielder has to position himself for the catch if he intends to catch the ball =2.84+3.47=6.30 sec.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes