Projectile motion and inital speed problem

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a projectile launched at an initial speed of 29.0 m/s at an angle of 65.0 degrees. The projectile is said to hit the ground after 8.00 seconds, and the questions focus on determining the height of the launch point relative to the landing point and the maximum height reached by the projectile.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of equations for horizontal and vertical motion, questioning the interpretation of the problem's wording and the use of initial conditions. Some express confusion about unit consistency and the significance of the launch height relative to the landing point.

Discussion Status

There is ongoing exploration of the problem, with participants offering insights into the equations used and the need for clarity in interpreting the problem statements. Some participants have shared their calculations and results, prompting further discussion on potential errors and misunderstandings.

Contextual Notes

Participants note the importance of expressing the launch height relative to the landing point and the maximum height relative to the launch point. There is also mention of the independence of horizontal and vertical motions in projectile motion analysis.

sunbunny
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Hey, I'm having trouble with this problem:

A projectile is fired with an initial speed of 29.0 at an angle of 65.0 degrees above the horizontal. The object hits the ground
8.00s later.

How much higher or lower is the launch point relative to the point where the projectile hits the ground?
Express a launch point that is lower than the point where the projectile hits the ground as a negative number.

To what maximum height above the launch point does the projectile rise?

So I'm pretty sure I've drawn my diagram right, and now I've tried to set up an equation where:

distance x = (vcos)t and

distance y = (vsine)t -.5gt^2

when i put in the values from the question i get the wrong answers. if anyone could point me in the right direction and tell me what I'm doing wrong that would be great.

Thanks!
 
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sunbunny said:
Hey, I'm having trouble with this problem:

A projectile is fired with an initial speed of 29.0 at an angle of 65.0 degrees above the horizontal. The object hits the ground
8.00s later.

How much higher or lower is the launch point relative to the point where the projectile hits the ground?
Express a launch point that is lower than the point where the projectile hits the ground as a negative number.

To what maximum height above the launch point does the projectile rise?

So I'm pretty sure I've drawn my diagram right, and now I've tried to set up an equation where:

distance x = (vcos)t and

distance y = (vsine)t -.5gt^2

when i put in the values from the question i get the wrong answers. if anyone could point me in the right direction and tell me what I'm doing wrong that would be great.

Thanks!
29 what? Are you doing the problem in the correct units? Maybe you are having difficulty because of the way the question is asked. Your equations are OK, but they assume x and y are zero at time zero. The first part of the question is worded such that it makes mores sense to use a y_o term that will make y = 0 at t = 8 sec. The later part is worded such that it makes more sense to do it as you have done. Make sure you are expressing the height of the launch point relative to the landing point in the first part, and the maximum height relative to the launch point for the second part.
 
oh sorry it's 29m/s
 
sunbunny said:
oh sorry it's 29m/s
OK.. your equations are good, so it's either a computation error or you are not interpreting their statements correctly.
 
i'm sorry to keep bothering you but honestly i just have no idea what's going on. If you could explain to me that would be really appreciated!

Thanks
 
sunbunny said:
i'm sorry to keep bothering you but honestly i just have no idea what's going on. If you could explain to me that would be really appreciated!

Thanks
I get that the launch point is 103.3m above the landing point. The maximum height is about 35.2m. That is what your equations should be giving you. What did you get? If it is different from mine, then we need to figure out why.
 
well my problem is that i set up the equations and then i don't know what to do with them. I usually just 'plug' in numbers but its not very useful for undersanding
 
sunbunny said:
well my problem is that i set up the equations and then i don't know what to do with them. I usually just 'plug' in numbers but its not very useful for undersanding
Right you are. In this problem you don't need the x equation at all. You just need to know that the y motion is independent of the x motion. You do need another equation of y motion for the second part. You can use

0 = Vosinθ - gt to find the time when the vertical velocity is zero (maximum height) and then use that time in your y equation to find y at that time.

You could also use V^2 = (Vosinθ)^2 - 2gh with V=0 to find the maximum height h.
 
well i know how to set up the equations, but then after that i just 'plug' in numbers but I'm not really understanding what is going on
 
  • #10
sunbunny said:
well i know how to set up the equations, but then after that i just 'plug' in numbers but I'm not really understanding what is going on
Practice will help. It can also be helpful to work with other people if you are trying to contribute and not just take what they do. It is one thing to watch other people do math or physics. It is quite something else to be doing it.

If you tell us more completely what you have tried, we can probably better understand what is wrong with your thinking. In this case you could have shown your calculations instead of just writing the equations, or at least posted the answers you got from them.
 
  • #11
okay thank you, that makes it more clear now of what i need to work and improve on and how i can show my steps more clear. Thank you so much for your help and your time, i really appreciated it!
 

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