How far does the catapult hurl the truck?

In summary: which comes out to 89.4 meters.5). find time of free fall from deck h=v*t, again from 2. t comes out to 1.8 seconds.6). the truck leaves the deck at v=296.157 m/s.
  • #1
honxkyu
4
0
heres the problem:
steam-powered catapult used to launch aircraft could "throw a nissan pick-up five miles(8045meters)"
catapult has the power to accelerate 30-ton aircraft from zero to 160mph(71.511m/s) in just 2.5 seconds
nissan pick up truck weighs 3500 pounds and the flight deck is 57 feet off the water
assuming no air resistance for the truck how do i find the actual distance the catapult hurls the truck?

ive found that
initial V = 0m/s
final V = 71.511m/s
acceleration = 28.604m/s^2

and I am stuck there :confused:
it also saids something about length of the deck but the value isn't given...

please give me a clue on what to do next
 
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  • #2
Since you've found out the acceleration, find the force it can give.(you know mass of aircraft.)This force can be used in throwing the truck!
 
  • #3
ergg
i found trucks acceleration as ~490m/s^2
and launch deck length as 89 meters..
it comes out that the truck travels .604 secs on the deck and after that, the truck is in free fall..
since the height of deck is 57 ft...its 17meters
and i calculated the time of free fall from 17 meters high so i got 1.8 secs..
the final vel when it just leaves the deck is 296.157 m/s
if i add 89meters and the length it travels horizontally during the free fall..its no where close to 5 miles..
am i doing something wrong??
 
  • #4
Hmm - this one is pretty confusing... it should be just distance = initial distance + velocity * time.
Can you write down the original question?
 
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  • #5
honxkyu said:
ergg
i found trucks acceleration as ~490m/s^2
and launch deck length as 89 meters..
it comes out that the truck travels .604 secs on the deck and after that, the truck is in free fall..
since the height of deck is 57 ft...its 17meters
and i calculated the time of free fall from 17 meters high so i got 1.8 secs..
the final vel when it just leaves the deck is 296.157 m/s
if i add 89meters and the length it travels horizontally during the free fall..its no where close to 5 miles..
am i doing something wrong??

how did u get the launch deck length??
 
  • #6
i wasnt really sure so i used the acceleration of aircraft and the time given
s=1/2 A t^s
i plugged in 28.604 for A and 2.5 for t
 
  • #7
The aircraft carrier USS constellation visited anchorage in 1986. During the visit, visitors were told that the steam powered catapult used to launch aircraft could "throw a nissan pick-up truck five miles" we will critique this boast

the following information is from the brochure which each constellation visitor received: each capapult has the power to accelerate 30ton aircraft from zero to 160mph in just 2.5 seconds

informations: nissan pick up is 3500 pounds and flight deck is 57 feet off the water

assume there is no air resistance for the truck..

this analysis should start with your deciding what factors would limit the distance the truck could move before hitting the water

convert all measurements into SI measurements
--------------------------------
this is the problem

futbol...if i just use distance = initial distance + velocity*time
i don't know the time or the velocity...and the truck is accelerated..

gah I am lost :bugeye:
 
  • #8
honxkyu said:
heres the problem:
steam-powered catapult used to launch aircraft could "throw a nissan pick-up five miles(8045meters)"
catapult has the power to accelerate 30-ton aircraft from zero to 160mph(71.511m/s) in just 2.5 seconds
nissan pick up truck weighs 3500 pounds and the flight deck is 57 feet off the water
assuming no air resistance for the truck how do i find the actual distance the catapult hurls the truck?

Emphasis is on the wrong word. "steam-powered catapult used to launch aircraft could throw a nissan pick-up five miles(8045meters)".

That doesn't mean the aircraft carrier and runway are currently configured to actually do that. The runway is configured as near to parallel with the ground (or water, in this case) as possible. Theoretically (not that it could happen in the real world), the runway could be angled to achieve a longer range for the truck. And it's longer than five miles - a little over 5 and a half miles.

So, yes, the actual distance an aircraft carrier designed for launching airplanes would chuck a truck is much, much shorter than the distance it could chuck a truck if an aircraft carrier could chuck trucks.
 
  • #9
I know the person who posted this has long since turned in that assignment, but nobody gave a numerical answer or good explanation, so for closure this is how i solved it. I have not checked my math and I paid no attention to sig. figs., but this is the basic process:
1). convert all given information into SI
2). find acceleration given to the given aircraft a=v/t. I got a=28.61 m/s^2
3). find the force supplied by the catapult f=ma. m is given mass of aircraft, a is acc. found in 2. I got f=778665 N
4). find length of carrier deck x=.5*a*t^2, a is acc. calculated in 2, t is given time. I got x=89.408 m
5). find acceleration of truck a=f/m, f is force calculated in 3, m is mass of truck. I got a=490.466
6). find velocity of truck at end of launch deck v^2=Vinitial^2+2ax, a is acc. calculated in 5, x is deck length calculated in 4. I got v=296.147.
7). find time for truck to fall from carrier deck to water x=.5*a*t^2, solve for t, a is acceleration due to gravity (9.80), x is distance from deck to water (57ft=17.3736m). I got t=1.88 s
8). find horizontal distance truck travels x=v*t, v is velcity calculated in 6, t is time calculated in 7. I got 557.6 m or 1829 ft.
Again, I have not rechecked my math, and i carried all decimal places in my calculator, but gave no attention to sig figs, but there it is, and I pretty sure that's how the problem should be solved.
 

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