The derivative as a rate of change

[XtremeChic04]

An object is dropped and hits the ground 6 seconds later. From what height, in feet, was it dropped?

I know:
a(t) = -32
v(t) = -32t +vo(initial velocity)
p(t) = -16t^2 + vot + po(initial position)

The answer in the back of the book is 576 ft. but I always end up with something around 1000. Anyhelp is greatly appreciated!

 

[Helwig]

Can you show me what you did..
Just write it out.

 

[XtremeChic04]

p(t) = -32(6)^2 + (0)(6) + po = 0
-1152 + po = 0
po = 1152

 

[soroban]

Hello, XtremeChic04!

You copied the function incorrectly . . .

p(t) = -32(6)² + (0)(6) + p_o = 0
.

 

[Helwig]

AHH HA!!! I see it... you made a tinny error...

p(t) = -32(6)^2 + (0)(6) + po = 0
-1152 + po = 0
po = 1152

You got this, but check again

p(t) = -16t^2 + vot + po(initial position) was what u needed to do... not -32

-16(6)^2 + (0)(6) = -576

then yes when -576 + po = 0 ; po = 576

 

[XtremeChic04]

oh man...i really do have difficulties with the easy stuff dont i? ha ha thank you very much! i guess i need to read better.