Angled Projectile Motion Question

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Homework Help Overview

The discussion revolves around a problem in projectile motion, specifically involving an object thrown at an angle of 22 degrees with a known horizontal distance of 38 meters. Participants explore the relationships between initial velocity, time of flight, and horizontal and vertical components of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss breaking down the initial velocity into horizontal and vertical components and the need for kinematic equations. Questions arise about the necessity of known velocities to solve for other variables, and there are attempts to derive time of flight in terms of initial velocity.

Discussion Status

The discussion has progressed with participants sharing various approaches and equations. Some have expressed confusion about specific variables and their meanings, while others have made attempts to connect horizontal and vertical motions. There is a recognition of the complexity of combining different components of velocity.

Contextual Notes

Participants mention the challenge of understanding the problem, particularly as it deviates from previous questions they found manageable. There is an acknowledgment of the potential for mistakes in calculations, especially when dealing with negative values in the context of projectile motion.

xxlvh
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1. An object was thrown up at an angle of 22o from the horizontal. If this object travels 38 m horizontally, what was the speed at which the object was thrown?

angle = 22o
dhorizontal = 38m
g = -9.81m/s2


I have absolutely no clue on how to even go about starting to solve this question...would anyone be able to give me a small hint perhaps on which formula to work with?
 
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Put the intial speed as V inclined at the angle of 22. Split that into vertical and horizontal components. (Note that the horizontal velocity is constant).
Then use your kinematic equations.
 
If I have the angle and split it into the x and y components, wouldn't I still need at least one velocity given to solve for any of the other sides?
 
Find the total time for the motion ([itex]s=ut+\frac{1}{2}at^2[/itex]) in terms of V.

Then the total horizontal distance travelled=Horizontal Speed * Total time for the motion
 
Hmm. The formulas I used were just slightly different, I've never seen that one before...what would the variables of "s" and "u" stand for?
 
xxlvh said:
Hmm. The formulas I used were just slightly different, I've never seen that one before...what would the variables of "s" and "u" stand for?

s=displacement
u=initial velocity
 
Okay, thank you.
And Vi would be...22?
Sorry for all the questions. I wasn't having any problems in the whole unit up until this last question!
 
xxlvh said:
Okay, thank you.
And Vi would be...22?
Sorry for all the questions. I wasn't having any problems in the whole unit up until this last question!

Vi=initial veritcal velocity=Vsin22.
 
Alright...I've gotten it to:

t = -2(Vsin22o) / -9.81m/s2
 
  • #10
xxlvh said:
Alright...I've gotten it to:

t = -2(Vsin22o) / -9.81m/s2

Good.

Now consider horizontal motion.

Horizontal Distance=Horizontal speed * Time


and you have t=(2Vsin22)/9.81 and the horizontal distance is 38
 
  • #11
Okay...So 38 m = V x (2vsin22/-9.81)

Are you able to combine the two V's, if one is horizontal and one is vertical?

When I tried that in the next step I have -372.78m=2v2sin 22o...when working through it though, it ends up with the root of a negative number so I'm assuming I made a mistake somwhere. =/
 
  • #12
xxlvh said:
Okay...So 38 m = V x (2vsin22/-9.81)

Are you able to combine the two V's, if one is horizontal and one is vertical?

When I tried that in the next step I have -372.78m=2v2sin 22o...when working through it though, it ends up with the root of a negative number so I'm assuming I made a mistake somwhere. =/

[tex]\frac{-2(Vsin22)}{ -9.81}[/tex]

minus signs cancel out.

and Horizontal velocity=Vcos22.
 
  • #13
Solving for V, I got 23 m/s. According to the book that is the correct answer!

This will add 3% onto my overall mark :) Thank you, I really owe you one!
 
  • #14
Once you understand how to do the problems is all that's needed.
 

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