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
a. What is the ball’s maximum height? 35Ft
b. When the ball hits the ground, how many seconds have passed? Show your work
Answers
Damon
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.
I will race you.
Damon
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
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
Hannah
The vertex is (5/4,35)
Damon
Beat ya :)
anyway part 2
16 t^2 - 40 t - 10 = 0
quadratic equation
a = 16
b = -40
c = -10
anyway part 2
16 t^2 - 40 t - 10 = 0
quadratic equation
a = 16
b = -40
c = -10
Damon
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
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
Damon
t = 2.73 for ground time
https://www.mathsisfun.com/quadratic-equation-solver.html
https://www.mathsisfun.com/quadratic-equation-solver.html