MEDAL
a bullet is fired horizontally at a velocity of 220 m/s. it strikes the ground at a distance of 511 m from where it was fired. how high above the ground was the bullet when it fired

- anonymous

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- NetherCreep333

Whats the question?

- anonymous

how high above the ground was the bullet when it fired

- anonymous

physics

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## More answers

- amistre64

start with maybe your linear equations of motion ....

- amistre64

if its fired straight out, there is no initial velocity up/down
and there is no acceleration acting on it from side to side since gravity pulls down, not out

- anonymous

d=vt

- amistre64

good, so t = d/v what was our time?

- anonymous

2.3 s

- amistre64

how far does something drop in 2.3 seconds?

- amistre64

1/2 gt^2 right?

- anonymous

is 2.3 correct

- amistre64

its good enough for me, but im not the one grading it. take it to however many decimals you want to

- anonymous

now do we use d=vit+1/2at^2

- amistre64

or just let it ride at 511/220

- amistre64

we can yes, vi=0 in this case

- amistre64

it more appropriate as: vi sin(theta) t
sin(0) = 0 as its shot horizontally

- anonymous

use vi sin(theta) t for the formula

- amistre64

that is a better generalization yes

- amistre64

\[h(t)=\frac12gt^2+v_i~t~\sin(\theta) \]

- amistre64

theta represent the angle of inclination from the horizontal

- anonymous

so what do i plug in into the equation

- amistre64

your values for gravity and time

- anonymous

what is the intial velocity
is it 0

- amistre64

since we are shot horizontally, then the degrees above horizontal is 0 ..
211 * sin(0) is your initial velocity in the up/down directions

- anonymous

so it is 0

- amistre64

yeah :)

- amistre64

so height = 1/2 gt^2

- anonymous

h(t) = 1/2(-9.8)(2.3)^2+0(2.3)+sin(0)

- amistre64

h(t) = 1/2(-9.8)(2.3)^2+211(2.3) sin(0)

- amistre64

h(t) = 1/2(-9.8)(2.3)^2+0
h(t) = 1/2(-9.8)(2.3)^2

- anonymous

why 211
should be 511

- amistre64

hmm, should be 220 i spose but i mistyped it since i didnt feel like scrolling all the way up to verify that it was correctly inputed.

- amistre64

but its immaterial, that term goes to 0

- amistre64

the full generalization equation for distance an object falls is:
\[h(t)=-\frac12gt^2+v_it~\sin(\theta )+h_i\]
h(2.3) = 0, we are on the ground
\[0=-\frac12g(2.3)^2+v_i(2.3)~\sin(0 )+h_i\]
\[\frac12g(2.3)^2=h_i\]

- anonymous

g=9.8

- amistre64

yes, or whatever the gravity of the planet your testing it on happens to be :)

- anonymous

25.92

- anonymous

i mean -25.92

- amistre64

and the dimensions is?

- anonymous

horizontal

- amistre64

no, initial height for this will be positive. you werent below ground to start with and it floated upwards ...

- amistre64

dimensions refered to feet, meters, yards, kilometers, etc ...

- anonymous

25.92 meters

- amistre64

good

- anonymous

is that the final answer

- amistre64

what is it we were looking for?

- anonymous

how high above the ground was the bullet when it fired

- amistre64

and did we find that?

- anonymous

i think so but not sure

- amistre64

h = 1/2 gt^2
so yeah, we found the h

- anonymous

can i ask another question

- amistre64

sure

- anonymous

A ball is thrown horizontally with an initial velocity of 20.0 meters per second from the top of a tower 60.0 high. What is the approximate total time required for the ball to reach the ground.

- amistre64

what was our generalization of finding height?

- anonymous

height equal 25.92

- amistre64

no, that was a specific case, not the general formula that i posted.

- anonymous

im on the next question

- amistre64

the great thing about generalizations, is that they work regardless of the question that you are on .. you simply reuse them with the new specifics.

- anonymous

d=vit+1/2at^2

- amistre64

almost, but then vi might confuse people if they dont understand that it pertains only to the up/down direction ..... hence the sin(theta)

- anonymous

so d=sin(theta) +1/2at^2

- amistre64

no, we dont eliminate the rest of it in favor of sin(theta)
the forces involved are this:
|dw:1446770512248:dw|
vi is important, and so is sin(theta). sin(theta) tells us how much of the initial velocity is imparted up/down

- amistre64

how many degrees from teh horizontal are we?

- anonymous

60

- anonymous

or 20

- amistre64

if its thrown horizontally, then we are zero degrees from the horizontal ...

- amistre64

"A ball is thrown horizontally"
it is zero degrees from the horizontal right?

- amistre64

h = vi t sin(0) + 1/2 at^2
h = 1/2 at^2
rearrange this to solve for t

- anonymous

t=1/2a^2

- amistre64

not quite ...
multiply by 2, divide by a, and sqrt

- amistre64

t = sqrt(2h/a)

- anonymous

what is the height

- amistre64

its defined in your problem ...

- anonymous

so 60

- amistre64

yes, is that meters?

- anonymous

t=sqrt(2(60))/9.8

- amistre64

sqrt it all. \[t=\sqrt{\frac{2(60)}{9.8}}\]

- anonymous

so 3.50

- amistre64

sounds reasonable to me

- anonymous

the correct answer is 3.50 sec

- amistre64

yes

- amistre64

good luck :)

- anonymous

one more question

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