Projectile Motion - object that is shot at an angle of 55 degrees

  • #1

Homework Statement


Projectile motion
I have an an object that is shot at an angle of 55 degree's. (the velocity listed next to it was 35 m/s, which was not listed as Vo but I think I have to assume it is?) It is launched from the ground. it is meant to land to hit a button that is elevated by 32 feet. we have to find the distance away from the button to launch the object in order to hit the button.

Homework Equations


Delta x= Vo * t
delta y = 1/2gt^2 + Vo * t

The Attempt at a Solution


I tried a bunch of things that did not help me come to any sort of conclusion about the distance (x)
 
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Answers and Replies

  • #2
gneill
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Please show us what you've tried.
 
  • #3
Charles Link
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For the ## x=v_x t ##, you need to have ## v_x=v_o \cos(55^{\circ}) ##. Do you see why? ## \\ ## For the y equation, you need ## y=-\frac{gt^2}{2}+v_{yi} t ## where ## v_{yi}=v_o \sin(55^{\circ}) ##. The acceleration due to gravity is downward=thereby the minus sign. ## \\ ## Meanwhile, an important part you are missing is to separate the initial velocity ## v_o ## at an angle of ## 55^{\circ} ## into horizontal and vertical components. I have shown you this result in the equations above. ## \\ ## Now that I got you started, can you figure out what to do to answer the question they are asking? Here's a hint: Can you solve for ## y ## as a function of ## x ##? ## \\ ## Presently, the equations you have go through the point ## x=0,y=0 ## at ## t=0 ##. What is ## x ## when ## y=32 ## ? You need to determine the equation that gives you ## y ## as a function of ## x ##, and then you can answer that question.
 
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  • #4
Please show us what you've tried.
sorry i didn't fill it out properly. i think i have a solution to the question now though, so i was wondeering if this thread could be deleted
 
  • #5
gneill
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sorry i didn't fill it out properly. i think i have a solution to the question now though, so i was wondeering if this thread could be deleted
After a thread has received replies, particularly where contributors have put in significant effort, our policy is to not remove the thread so that other members can benefit from the information.
 
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  • #6
Ray Vickson
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Homework Statement


Projectile motion
I have an an object that is shot at an angle of 55 degree's. (the velocity listed next to it was 35 m/s, which was not listed as Vo but I think I have to assume it is?) It is launched from the ground. it is meant to land to hit a button that is elevated by 32 feet. we have to find the distance away from the button to launch the object in order to hit the button.

Homework Equations


Delta x= Vo * t
delta y = 1/2gt^2 + Vo * t

The Attempt at a Solution


I tried a bunch of things that did not help me come to any sort of conclusion about the distance (x)
Be very careful. If ##y## measures height above the ground, your formula ##\Delta y = v_0 t + \frac{1}{2} g t^2## will have the object accelerating rapidly upward, until it goes into outer space (because, conventionally, ##g## is a positive constant).
 
  • #7
Be very careful. If ##y## measures height above the ground, your formula ##\Delta y = v_0 t + \frac{1}{2} g t^2## will have the object accelerating rapidly upward, until it goes into outer space (because, conventionally, ##g## is a positive constant).
:eek:How do I change the equation to keep it from accelerating into outer space?
 
  • #8
Charles Link
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:eek:How do I change the equation to keep it from accelerating into outer space?
You put a minus sign in front of the ## \frac{g \, t^2}{2} ## term. ## g=+9.8 ## m/sec^2, but ## a=-9.8 ## m/sec^2. ## \\ ## Otherwise, your equation for ## y ## has the form ## y=At^2+Bt ## with ## A>0 ## , so that ## y ## will get increasingly larger with time ## t ##.
 
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  • #9
You put a minus sign in front of the ## \frac{g \, t^2}{2} ## term. ## g=+9.8 ## m/sec^2, but ## a=-9.8 ## m/sec^2. ## \\ ## Otherwise, your equation for ## y ## has the form ## y=At^2+Bt ## with ## A>0 ## , so that ## y ## will get increasingly larger with time ## t ##.
You are a life, saver thank you very much for your explanations
 
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