A object falls at 16t^2 feet in t seconds you drop a rock from a bridge that is 906 what is the hight in feet after the rock is droped

Respuesta :

Answer:

824 feet

Step-by-step explanation:

Given

[tex]h(t) = 16t^2[/tex]

[tex]H = 906[/tex] -- Initial Height

Required

Calculate the height after 2 seconds --- missing from the question

First, we calculate the distance moved in 2 seconds

[tex]h(t) = 16t^2[/tex]

[tex]h(2) = 16*2^2[/tex]

[tex]h(2) = 16*4[/tex]

[tex]h(2) = 64[/tex]

This means that the object covered 64feet

The object current height is then calculated as:

[tex]Height = H - h(t)[/tex]

t = 2. So, the equation becomes

[tex]Height = H - h(2)[/tex]

Substitute values for H and h(2)

[tex]Height = 906- 64[/tex]

[tex]Height = 824[/tex]