# If a ball reached its maximum height of 106 feet at time=2.5, what is the height, in feet, of the ball at time t=1?At time t=0, a ball was thrown upward from an initial height of 6 feet. Until the...

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### 1 Answer

h(t)= c- (d-4t)^2

First we know when t= 0==> h= 6

`==gt 6= c-(d-0)^2= c-d^2 `

`==gt c-d^2= 6` ..............(1)

When h= 106, ==> t= 2.5

`==gt 106= c- (d-4*2.5)^2 `

`==gt 106= c-(d-10)^2 `

`==gt 106= c- (d^2 -20d+100) `

`==gt 106= c-d^2 + 20d -100`

`` But `c-d^2 = 6`

`` `==gt 106= 6 + 20d -100 `

`==gt 20d= 200 `

`==gt d= 200/20= 10 `

`==gt c= d^2+6 `

`==gt c= 106 `

`==gt h(t)= 106- (10-4t)^2`

`` We need to find h when t= 1

`==gt h(1)=106 - (10-4)^2 `

`==gt h(1)= 106- 36= 70`

`` **Then the height when (t= 1) is 70 ft.**