Question

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

Answer 1

Whats the question?

Answer 2

how high above the ground was the bullet when it fired

Answer 3

physics

Answer 4

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

Answer 5

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

Answer 6

d=vt

Answer 7

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

Answer 8

2.3 s

Answer 9

how far does something drop in 2.3 seconds?

Answer 10

1/2 gt^2 right?

Answer 11

is 2.3 correct

Answer 12

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

Answer 13

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

Answer 14

or just let it ride at 511/220

Answer 15

we can yes, vi=0 in this case

Answer 16

it more appropriate as: vi sin(theta) t

sin(0) = 0 as its shot horizontally

Answer 17

use vi sin(theta) t for the formula

Answer 18

that is a better generalization yes

Answer 19

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

Answer 20

theta represent the angle of inclination from the horizontal

Answer 21

so what do i plug in into the equation

Answer 22

your values for gravity and time

Answer 23

what is the intial velocity
is it 0

Answer 24

since we are shot horizontally, then the degrees above horizontal is 0 ..

211 * sin(0) is your initial velocity in the up/down directions

Answer 25

so it is 0

Answer 26

yeah :)

Answer 27

so height = 1/2 gt^2

Answer 28

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

Answer 29

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

Answer 30

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

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

Answer 31

why 211
should be 511

Answer 32

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.

Answer 33

but its immaterial, that term goes to 0

Answer 34

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
\[0=-\frac12g(2.3)^2+v_i(2.3)~\sin(0 )+h_i\]
\[\frac12g(2.3)^2=h_i\]

Answer 35

g=9.8

Answer 36

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

Answer 37

25.92

Answer 38

i mean -25.92

Answer 39

and the dimensions is?

Answer 40

horizontal

Answer 41

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

Answer 42

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

Answer 43

25.92 meters

Answer 44

good

Answer 45

is that the final answer

Answer 46

what is it we were looking for?

Answer 47

how high above the ground was the bullet when it fired

Answer 48

and did we find that?

Answer 49

i think so but not sure

Answer 50

h = 1/2 gt^2

so yeah, we found the h

Answer 51

can i ask another question

Answer 52

sure

Answer 53

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.

Answer 54

what was our generalization of finding height?

Answer 55

height equal 25.92

Answer 56

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

Answer 57

im on the next question

Answer 58

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.

Answer 59

d=vit+1/2at^2

Answer 60

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

Answer 61

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

Answer 62

no, we dont eliminate the rest of it in favor of sin(theta)

the forces involved are this:



vi is important, and so is sin(theta). sin(theta) tells us how much of the initial velocity is imparted up/down