Basic Projectile help needed

  • Thread starter Aldy
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  • #1
Aldy
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Homework Statement


a ball is launched vertically from a spring loaded pipe at ground level, and it lands in 4.7 seconds. We need to figure out the launch velocity and how far the ball will land if fired at 45 deg angle.

Homework Equations


I am using the following formulas, not sure if they are the correct ones to use.
V= -1/2gt
dy= Voyt + 1/2ayt (squared)
dx = Voxt + 1/2axt(squared)


The Attempt at a Solution


I calculated the launch velocity to be 23m/s
however I am having trouble calculating how far the ball lands if launched at 45deg angle.
The answer should be between 50-60 meters but I get an answer of 155.

Could someone please help me figure this out, please show work so I can understand. Thanks

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Alucinor
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Erasing this for confusing language and a mistake:redface:, sorry
 
Last edited:
  • #3
Aldy
Gold Member
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Erasing due to incorrect info
 
Last edited:
  • #4
Alucinor
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How about this, I think I need to proofread my posts a little better, I actually might have made a mistake in that one :redface:

Do you agree that, for a projectile launched at an angle theta, you can decompose an initial velocity as such:
[tex]v_{0x}=v_0cos(\theta)[/tex]
[tex]v_{0y}=v_0sin(\theta)[/tex]

Well, then, from the equation to follow we can calculate the initial y velocity (because initial y location is known, it is 0, y acceleration is known, it is g, and t is known, it is 4.7s, and its final location is known, which is when y is equal to 0 again, you should graph this function, it is a parabola)
[tex]y=y_0+v_{0y}t+\frac{1}{2}a_yt^2[/tex]
Then, with that figured out, we can get the total velocity using and then the x velocity in a few steps using those first two equations I gave you.

After solving for the initial x direction velocity, you can use the equation
[tex]x=x_0+v_{0x}t+\frac{1}{2}a_xt^2[/tex]
to find the total x displacement. We also know the initial x location, 0, and the x acceleration, 0 (because we have no air resistance), and now we know the initial x velocity from the work before.

Does that help?

It is important to realize that velocity is a vector, it has magnitude and direction. In a kinematics problem we can treat the x and y directions as independent and decompose a velocity vector into its constituent x and y components with the first two equations I gave you, and then use them in independent kinematic equations.
 
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  • #5
Aldy
Gold Member
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Thank you, that is very helfpul.:smile:
 
  • #6
Alucinor
29
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Please delete that quote so we don't confuse anyone who might end up here
 

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