Question

A ball is thrown into the air with an upward velocity of 40 feet per second. Its height, h, in feet after t seconds is given by the function h(t)=-16t^2+40t+10.
a. What is the ball’s maximum height? 35Ft
b. When the ball hits the ground, how many seconds have passed? Show your work

Answer 1

I assume you do not do calculus so find the vertex of that parabola. That is just like the last one we did :)
I will race you.

Answer 2

16 t^2 - 40 t - 10 = -h

divide by 16 first to get 1 as coef of t^2
t^2 - 2.5 t - .625 = -h/16
t^2 - 2.5 t = -h/16 + .625
half of 2.5 then square
get 1.5625, add to both sides
t^2 - 2.5 t + 1.5625 =-h/16 + 1.5625
(t-1.25)^2 = -(1/16)(h-25)
vertex (top of the parabola) at
h = 25
t = 1.25 seconds
now solve for t when h = 0 .
Use the + answer

Answer 3

The vertex is (5/4,35)

Answer 4

Beat ya :)
anyway part 2
16 t^2 - 40 t - 10 = 0
quadratic equation
a = 16
b = -40
c = -10

Answer 5

we agree about t, but not about h
CHECK
16 t^2 - 40 t - 10 = -h
16(1.25)^2 - 40(1.25) - 10 = ?
25 - 50 -10
-35 = -h , you are right
I forgot to add 10 on the right

Answer 6

t = 2.73 for ground time
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