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Projectile Motion

  • Thread starter MC Escher
  • Start date
  • #1
13
0

Homework Statement


An object was launched from the top of a cliff at some angle with an acceleration of 50 ft/sec that overcame gravity. The fuel tank ran out after 2 seconds of flight and gravity took over from that point forward. The object reached a maximum height of 310 ft, was in the air for a total of 9.16 seconds and traveled a horizontal distance of 383 ft.
Find:
Initial Launch Angle
Height of the cliff
Initial vertical velocity
Initial horizontal velocity


Homework Equations



s(t)=1/2at^2+vt+s
d=1/2at^2
Vf=Vi+at

The Attempt at a Solution


I think I may have found the launch angle by using the 1/2at^2 equation and plugging in 383=1/2ax(9.16)^2 and 310=1/2ay(9.16)^2. I found the angle to be 39.36, but I am unsure if it is correct, after that I am stuck, help is greatly appreciated.
 

Answers and Replies

  • #2
13
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anyone?.....
 
  • #3
960
0
lets see, the first thing to sieze on is the horizontal displacement:

if its 383' and t=9.16 we know, Vx=41.8'/s (ave velocity)

moreover we know that the force/m applied was 5*g, and the duration 2 seconds.

The initial x velocity of rocket 0.

so for the x component we have a displacement of two components:

that covered during acceleration phase and that while coasting here a is the x component of acceleration

383=(1/2at^2)+ at(t') where t+t'=9.16 and since t=2, t'=7.16, hence a

383=.5a*(2^2)+a*2(7.16) we have 2a+14.32a=383 or 16.32a=383 a=23.47

so atotal/ax=5*32.2/23.47=6.86 ie arccos=1/6.86=81 degrees. Sure you can manage from here. This any help?
 
Last edited:
  • #4
13
0
thanks, having the proper angle is a big help...im having a tough time with the height though...
 
  • #5
If its launched at an angle theta, then for the first two seconds, its also accelerating at 50 ft/sec^2. So, resolving the acceleration into horizontal and perpendicular components, you can find the initial speed in the x (horizontal) direction.

Now, in the vertical direction, there are two forces/accelerations acting. Their vector sum will give you the net acceleration, and this (over a period of 2 seconds) will give you the initial speed in the y direction.

Since you know what the max height is, you can use the above to get an equation in theta/time. And using the horizontal distance you get another equation in theta/time. Two equations and two variables, you get your answer.
 
  • #6
13
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sorry, but I don't follow.....I have it set up now as 383=1/2ax(9.16)^2 and 310=1/2ay(9.16)^2, is that what you are saying?
 
  • #7
960
0
well its a bear of a problem. ugliest w/o air resistance I've seen in a while. But I can't do all the work. So a hint will have to suffice:
parce it out for the two phases, with ay =sin81(5*g) at 2 seconds for phase 1, then it becomes purely kinematic. so you need to keep track of the altitude above the cliff at 2 seconds, and the fact during purely ballistic phase it starts at that altitude and lands below both the launch site and the time where it became purely ballistic.
 
  • #8
13
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thanks, im going to work on it for a little while longer and then call it a night...lol
 
  • #9
960
0
have fun, its pugugly but doable. You'll be a manlier man for having done so?? dont forget gravity during both phases!!!!
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,808
933

Homework Statement


An object was launched from the top of a cliff at some angle with an acceleration of 50 ft/sec that overcame gravity. The fuel tank ran out after 2 seconds of flight and gravity took over from that point forward. The object reached a maximum height of 310 ft, was in the air for a total of 9.16 seconds and traveled a horizontal distance of 383 ft.
Find:
Initial Launch Angle
Height of the cliff
Initial vertical velocity
Initial horizontal velocity


Homework Equations



s(t)=1/2at^2+vt+s
d=1/2at^2
Vf=Vi+at

The Attempt at a Solution


I think I may have found the launch angle by using the 1/2at^2 equation and plugging in 383=1/2ax(9.16)^2 and 310=1/2ay(9.16)^2. I found the angle to be 39.36, but I am unsure if it is correct, after that I am stuck, help is greatly appreciated.
Exactly HOW did you find the launch angle? Please show your work.
The maximum height is reached in 9.16 seconds but NOT its full horizontal distance!
 
  • #11
13
0
any more help with this? I just keep getting stuck...
 
  • #12
960
0
please note mistake above from english units--I put in 5*g for acceleration when it was 50'/sec^2. My bad. So much more used to Mks. The approach is the same. the angle should be 50/23.6=1/arccos=61.8 degrees, sorry.
 
  • #13
hotvette
Homework Helper
990
3
Also, if you haven't done it already, it REALLY helps to draw a picture labeling all relevant information. It helps relate the equations to the physical situation and keep signs straight. Free Body Diagram........
 
  • #14
13
0
I finally got the problem finished earlier today, I will post my results up later...it was pretty tough...thanks for help
 
  • #15
13
0
please note mistake above from english units--I put in 5*g for acceleration when it was 50'/sec^2. My bad. So much more used to Mks. The approach is the same. the angle should be 50/23.6=1/arccos=61.8 degrees, sorry.
yea, I found that to be the angle also, I finished up the problem earlier, I will post up my results soon, thanks for help...
 
  • #16
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
18
I've merged your two threads.
 
  • #17
960
0
thanks, i looked at this tonite with all the interdigitatated posts and wondered whether it was me or the Twilight zone that PF had entered.
 

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