Expert: Sherman D. - 10/24/2008

Question
If a ball is thrown upward from a building 30m tall and the ball has a vertical velocity of 25 m/s, then its approximate height above the ground t seconds later is given by h(t)= 30 + 25t - 5t^2.

A)after how many seconds does the ball hit the ground?
B)what is the domain of h?
C)how high does the ball go?

For part C my answer was 61.25 meters high. Is that correct? And I really need help on finding the domain.

Thanks ; ]

Answer
h(t) = -5t^2 + 25t + 30

h(t) = -5(t^2 - 5t - 6)
h(t) = -5(t - 6)(t + 1)

a.)
t = 6 seconds

b.)
-1 < x < 6

c.)
t = -25/(2(-5))
t = -25/-10
t = 2.5

h(5/2) = -5(5/2)^2 + 25(5/2) + 30
h(5/2) = -5(25/4) + (125/2) + 30
h(5/2) = (-125/4) + (125/2) + 30
h(5/2) = (-125/4) + (250/4) + (120/4)
h(5/2) = (-125 + 250 + 120)/4
h(5/2) = (125 + 120)/4
h(5/2) = (245/4)
h(5/2) = 61.25

The max height is 61.25 meters