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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?

 Mar 2, 2021

Best Answer 

 #1
avatar+31281 
+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 ! cheeky

 Mar 2, 2021
 #1
avatar+31281 
+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 ! cheeky

ElectricPavlov Mar 2, 2021

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