Projectile Motion given a vertical distance and angle at that point

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Homework Help Overview

The discussion revolves around a projectile motion problem where the vertical distance is known, and the angle of the velocity vector at that point is given as 45 degrees. Participants are exploring the implications of this angle on the horizontal and vertical components of velocity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the horizontal and vertical components of velocity, noting that at 45 degrees, these components are equal. Questions arise regarding the acceleration in both the x and y directions and how to apply the equations of motion to the given conditions.

Discussion Status

The discussion has seen some participants express uncertainty about how to proceed with the problem, while others have pointed out key aspects of the motion, such as the lack of acceleration in the x-direction and the presence of acceleration in the y-direction. One participant has indicated a breakthrough in understanding, suggesting that some productive direction has been achieved.

Contextual Notes

Participants are working within the constraints of the problem, which includes specific values for vertical distance and the angle of the velocity vector. There is an acknowledgment of the need to utilize the "Famous Five" equations for uniform accelerated motion, but some participants are struggling to connect these equations to the information provided.

whoareyou
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Homework Statement


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Homework Equations


Famous Five equations for uniform accelerated motion

The Attempt at a Solution


To be honest, I have absolutely no idea where to start. The only thing that I notice is that at the point where you're given the velocity vector (when the projectile is 10m above the ground) it makes an angle of 45 degrees, which means that both vx and vy are equal. I don't really know where to go from there.
 
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whoareyou said:
To be honest, I have absolutely no idea where to start. The only thing that I notice is that at the point where you're given the velocity vector (when the projectile is 10m above the ground) it makes an angle of 45 degrees, which means that both vx and vy are equal. I don't really know where to go from there.

I'm glad you recognize Vx and Vy are equal. That is key to being able to solve the problem. :smile:

Since you know Vx and Vy are equal, think about what else you know.
What is the acceleration in the x-direction?
What about in the y-direction?

Knowing the starting position (just clearing the wall) can you use any of those "Famous Five" with Vx and Vy and your known accelerations to set up an expression that will let you solve for more information?
 
There is no acceleration in the x direction, delta d x = vx delta t.
There is acceleration in the y direction, delta d = viy delta t + 1/2 g delta t squared

delta d y = delta d x x time + 1/2 g t squared --> i end up with -18 = 4.9t^2 which can't be done.
 
NVM! got it YAY!
 
whoareyou said:
NVM! got it YAY!

Alright then. Glad to hear it. :approve:
 

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