Imagine that someone dropped a firecracker off the roof of a building and heard it explode exactly 10 s later. Ignoring air friction, taking g= 9.81 and using 330 m/s as the speed of sound, calculate how far the cherry bomb had fallen at the very moment it blew up.
2006-07-30 07:23:34 · 9 answers · asked by MelBelle 2 in Science & Mathematics ➔ Physics
dont say anything unless you can really help me, I appriciate the funny remarks...but they keep getting my hopes up when I'm emailed "you have an answer" when I really dont
2006-07-30 07:34:01 · update #1
oh and this is all the variables that were given to me, if there were more, I wouldnt ask. Like I wanna know what the the time it took in at least one direction. Its soooo frustrating...ya feel me?
2006-07-30 07:35:48 · update #2
do you think the problem meant, that the dude on the roof heard the sound 10 s after it exploded or 10 s after he dropped it?
2006-07-30 07:37:57 · update #3
ha ha! Ariel is the winner! Hey, where were you guys like 4 days ago when I had a Buoyancy problem. NOBODY jumped on it. I just answered it yesterday. thanx for this one though. yay!
2006-07-30 08:28:01 · update #4
here is your answer:
the guy dropped it from the roof, it fell through 383.3m and then exploded.
additional info:it took 8.84secs to fall through 383.3m and then exploded. it took another 1.16 sec for the sound to reach the guy on the roof. so totally it took 10 secs for the guy to hear the explosion!
2006-07-30 07:46:47 · answer #1 · answered by Ariel 2 · 1⤊ 2⤋
Too many variables,, depends on the length of the fuse , hight of the building aerodynamics of the firecracker itself, but most likely it only fell about 14 feet,, then hit the sidewalk and bounced a couple of times before it blew up and scared the hell out of the nice old lady that was satnding there waiting for the bus at the time. On average that would take about 6.3 seconds.
2006-07-30 14:29:24 · answer #2 · answered by sparky_the_perv 3 · 0⤊ 0⤋
383.304 metres.
I started to put in some numbers. I took the time taken for fall as 8.84s and worked out the distance,
s=1/2.g.t^2 = 1/2 . 9.81 . 8.84^2 = 383.304m
Now sound travels at 330m/s. At that speed it would take,
t=s/v = 383.304 / 330 = 1.6s
So, total time = 8.84 + 1.6 = 10s
Note: to be fair Ariel managed to answer you before me. Well done dude! It took me a few minutes (and a number of scribbled pages) to work it out. I also thinkk there is some sort of connection between the equations, but I hav'nt figured it out yet
2006-07-30 15:19:32 · answer #3 · answered by alexsopos 2 · 0⤊ 0⤋
I don't think you can solve it as you have stated it. You can find the distance the sound traveled by
distance1= time1(330)m/s
and you can find the distance it falls by
distance2 = (1/2)g(time2)squared
and then time1 + time2 must equal ten seconds. Set the two distances equal, since they are the same distance. But this is two equations in two unknowns (two different portions of time.)
If it is known that the explosion occurs ten seconds after you drop it, then use the second equation to find the distance, and then use the first equation to find the time the sound requires to travel back that distance to you (AFTER the ten-second elapsed time of the explosion.)
Hope this helps! Good luck!
2006-07-30 14:53:27 · answer #4 · answered by cdf-rom 7 · 0⤊ 0⤋
Congatulations Ariel.
2006-07-30 15:21:38 · answer #5 · answered by heinlein 4 · 0⤊ 0⤋
distance= 10s x 330m/s =3,300 m
2006-07-30 14:35:03 · answer #6 · answered by idiotsavant 1 · 0⤊ 0⤋
way below the roof. lol
2006-07-30 14:31:42 · answer #7 · answered by R_H 1 · 0⤊ 0⤋
about the floor somwhere
2006-07-30 14:27:45 · answer #8 · answered by Dee 4 · 0⤊ 0⤋
what is the weight of the fire cracker
2006-07-30 14:33:13 · answer #9 · answered by maruawe@sbcglobal.net 2 · 0⤊ 0⤋