A rock is thrown into the air. It's height is given by the functionh(t) = -16t^2+ 64t + 17 where his measured in feet and t in seconds.

What is the height of the rock after 1 second?

At what time is the rock the highest?

t=

What is the highest the rock goes?

When does the rock hit the ground?

Guest Mar 2, 2021

#1**+2 **

h(t) = -16t^2+ 64t + 17

put in t = 1 and calculate the height at 1 second

Highest point occurs a t t = -b/2a where b = 64 and a = -16 use this value of 't' in the equation to calculate the highest point

When the rock hits the ground h = 0

sooooo 0 = -16t^2 + 64 t + 17 use Quadratic Formula to find 't' values (throw out the negative 't' value)

quadratic formula a = -16 b = 64 c = 17

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\) plug and play !

ElectricPavlov Mar 2, 2021

#1**+2 **

Best Answer

h(t) = -16t^2+ 64t + 17

put in t = 1 and calculate the height at 1 second

Highest point occurs a t t = -b/2a where b = 64 and a = -16 use this value of 't' in the equation to calculate the highest point

When the rock hits the ground h = 0

sooooo 0 = -16t^2 + 64 t + 17 use Quadratic Formula to find 't' values (throw out the negative 't' value)

quadratic formula a = -16 b = 64 c = 17

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\) plug and play !

ElectricPavlov Mar 2, 2021