Question

Your friend competes in a high jump competition at a track meet. The function h= -16t^2 + 18t models the height h (in feet) of your friend after t seconds. After how many seconds is your friend at a height of 4 feet? After how many seconds does your friend land on the ground?

Answer 1

I have attached all my work below. I hope it all makes sense. If not, or if you still have questions feel free to comment below! :)

Answer 2

Time when the height is 4 feet is given by

`t= 0.82` sec or
`t=0.30`sec

and time taken to reach the ground is 1.125 sec

Explanation:

It is given that the height is given by

`h= -16t^2 +18t`

now we need to find the time t when height h= 4 feet , so we plug h= 4 and solve for t, so we have

`4=-16t^2 +18t`

`16t^2-18t+4=0` (adding
`16t^2` and subtracting
`18t` to both sides)

now we can use quadratic formula

`t=(-b\pm√(b^2-4ac) )/(2a)`

we compare
`16t^2-18t+4=0` with
`at^2+bt+c=0`

we have
`a=16 , b=-18, c=4`

Now we have

`t=(-(-18)\pm√((-18)^2-4(16)(4)) )/(2(16))`

`t=(18\pm√(68) )/(32)`

`t=(18\pm8.246 )/(32)`

`t=(18+8.246)/(32)` or
`t=(18-8.246)/(32)`

so the time when the height is 4 feet is given by

`t= 0.82` sec or
`t=0.30`sec

Now height is 0 on the ground, so we plug h=0 to find t

`0=-16t^2 +18t`

`16t^2 -18t=0`( we bring all the terms to left side)

`2t(8t-9)=0`

`t=0` or
`8t-9=0` ( we equate each factor to 0)

`t=0` or
`t=(9)/(8) = 1.125`

t=0 is the initial time

hence time taken to reach the ground is 1.125 sec