#1

Ok i dont know how to do this.

On a day when the speed of sound in air is 330 m/s, you shoot a rifle at a distant metal target. The speed of the bullet is 500 m/s. You hear the sound of the bullet striking the target just 5 seconds after firing the gun. Calculate the distance from you to the target.

The teacher gave us the answer (990m) but i need to know how to get that answer.

On a day when the speed of sound in air is 330 m/s, you shoot a rifle at a distant metal target. The speed of the bullet is 500 m/s. You hear the sound of the bullet striking the target just 5 seconds after firing the gun. Calculate the distance from you to the target.

The teacher gave us the answer (990m) but i need to know how to get that answer.

#2

but i need to know how to get that answer.

The key to success here is to work it out.

I know how but this is not my work to do

#3

Should be fairly straightforward...

Damn, I haven't had physics in two years but I'ma give it a crack anyway.

Damn, I haven't had physics in two years but I'ma give it a crack anyway.

#4

what.???

i'm stupid and so are a few other people on here

i'm stupid and so are a few other people on here

#5

v=d/t

v= 330m/s

t= 5s

d=?

v=d/t becomes d=v*t

figure the rest out yourself.

v= 330m/s

t= 5s

d=?

v=d/t becomes d=v*t

figure the rest out yourself.

*Last edited by J.A.M at Oct 12, 2007,*

#6

Well you obviously multiply the 330 m/s by 3 to get the answer, which is 990m.

Duh.

Duh.

*Last edited by X~THOR~X at Oct 12, 2007,*

#7

Hang on,

is your teachers answer right? or am i wrong?

is your teachers answer right? or am i wrong?

#8

It seems like you need the time traveled each way, like that's the key to this.

#9

well the answer should be 2500m if you use the whole velocity *time= distance, but i have a feeling that this problem is a little more complicated than it seems...wat kind of physics is this? is it basic conceptual or is it college calculus based?

#10

I can get a distance of 1km. That's close to 990m, right?

#11

Hang on,

is your teachers answer right? or am i wrong?

I am wondering the same thing...

EDIT: actually no, 990 is correct... small miscalculation

#12

OK, here's what I've got, which actually takes no physics knowledge.

The ratio of the speed of sound to the speed of the bullet = .66; that is, in the time it takes the bullet to go 1m, the sound travels .66m.

Retardedly, for this to work out, you must move 10m closer to the target in those 5s.

In 2s, the bullet moves 1000m or 1km. In the other 3s, the sound moves 990m, which is your answer.

The ratio of the speed of sound to the speed of the bullet = .66; that is, in the time it takes the bullet to go 1m, the sound travels .66m.

Retardedly, for this to work out, you must move 10m closer to the target in those 5s.

In 2s, the bullet moves 1000m or 1km. In the other 3s, the sound moves 990m, which is your answer.

#13

well the answer should be 2500m if you use the whole velocity *time= distance, but i have a feeling that this problem is a little more complicated than it seems...wat kind of physics is this? is it basic conceptual or is it college calculus based?

You have 2 parts to consider. First, the time taken for the bullet to hit the target. Second, the time taken for the sound of THAT impact to reach you.

If I'd paid more attention in maths(or if I'd been in a maths class more recently than 9-10 years ago), it'd be simple.

#14

Ok i dont know how to do this.

On a day when the speed of sound in air is 330 m/s, you shoot a rifle at a distant metal target. The speed of the bullet is 500 m/s. You hear the sound of the bullet striking the target just 5 seconds after firing the gun. Calculate the distance from you to the target.

The teacher gave us the answer (990m) but i need to know how to get that answer.

Call time taken for bullet to get to target t1

Call time taken for sound to get from target to observer t2

Call distance from target to observer x

Total time t1 + t2 = 5s ... Call this equation (1)

t1 = x / 500 ... Call this equation (2)

t2 = x / 330 ... Call this equation (3)

Therefore, subbing (2) and (3) into (1) gives:

(x / 500) + (x / 330) = 5

Cross multiply to give:

830x / 165000 = 5

Rearrange:

x = 994.0 m

#15

EDIT- ^ yeh that's what I wanted to do but didn't know how

*Last edited by rankine at Oct 12, 2007,*

#16

Call time taken for bullet to get to target t1

Call time taken for sound to get from target to observer t2

Call distance from target to observer x

Total time t1 + t2 = 5s ... Call this equation (1)

t1 = x / 500 ... Call this equation (2)

t2 = x / 330 ... Call this equation (3)

Therefore, subbing (2) and (3) into (1) gives:

(x / 500) + (x / 330) = 5

Cross multiply to give:

830x / 165000 = 5

Rearrange:

x = 994.0 m

+1

I'm depressed at how many of you failed to answer this incredibly basic problem.

#17

+1

I'm depressed at how many of you failed to answer this incredibly basic problem.

Why should we answer it for him, though? Since it has been answered for him in detail, it renders the exercise valueless. Not that it was overly valuable to begin with.

#18

x = time for bullet to hit target

y = time for sound to return to shooter

x + y = 5

x = 5 - y

330y = 500x

330y = 500(5 - y)

y = 250/83

x = 5 - (250/83)

(Skipped the Algebra)

d = 330*(250/83) = 993

If you round off 250/83 to 3 it is 990m.

EDIT: Beat me to it.

y = time for sound to return to shooter

x + y = 5

x = 5 - y

330y = 500x

330y = 500(5 - y)

y = 250/83

x = 5 - (250/83)

(Skipped the Algebra)

d = 330*(250/83) = 993

If you round off 250/83 to 3 it is 990m.

EDIT: Beat me to it.

#19

Why should we answer it for him, though? Since it has been answered for him in detail, it renders the exercise valueless. Not that it was overly valuable to begin with.

I never said you should, only that most people who tried or hinted failed miserably.

#20

I never said you should, only that most people who tried or hinted failed miserably.

Like me! Yay!!!