vertical motion

G

Guest

Guest
you are standing on a brindge over a creek holding a stone 20 feet above the water
you selease the stone ...How long will it take the stone to hit the water?

I have h= -16 t (squared) + s
I inputed my data to get
h = -16 t(squared) + 20
then got -20/-16 = -16 t (sqaured)/-16
to get a 11/4 = t (squared)
t= about 1 second?

am I anywhere close???
:?
 
Calling upwards + and downwards -

You are at a distance +20 and want to find when the rock hits 0.

You had some of it right...

\(\displaystyle 16t^2-20=0\)

Dividing through by 4...

\(\displaystyle 4t^2-5=0\)

Solving for t and disregarding the negative answer,

\(\displaystyle t=\sqrt{\frac{5}{4}}\Rightarrow t=\frac{\sqrt{5}}{2}\)
 
still lost as what amount of time would that be???

I have it figured out to about 1 sec but know that's too soon??? :? :?
 
Re: still lost as what amount of time would that be???

4 little piggies mom said:
I have it figured out to about 1 sec but know that's too soon??? :? :?

Let's use some common sense. The force of gravity will cause the rock to drop 16 t<SUP>2</SUP> feet in t seconds. In ONE second, then, the rock will fall 16(1)<SUP>2</SUP> feet, or 16 feet. Your rock needs to fall just 20 feet, so the time should be just a bit over 1 second. Look at the answer you were given by Miniputtz.....
 
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