Supposedly easy physics problem =( [projectile]

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SUMMARY

The discussion revolves around solving a projectile motion problem involving an initial velocity of 4 m/s at a 60-degree angle, targeting a distance of 1.0 m horizontally. Key calculations include determining the y-coordinate of the target, which was found to be approximately 3.2 meters, and the time to hit the target, which was debated between 0.25 seconds and 0.5 seconds. The final vertical velocity was calculated to be -1.44 m/s, and the overall speed before impact was derived using the Pythagorean theorem, resulting in a magnitude of 3.99 m/s at an angle of 59.97 degrees. The discussion highlights common misconceptions and emphasizes the importance of using correct initial conditions and equations in projectile motion.

PREREQUISITES
  • Understanding of basic kinematics and projectile motion.
  • Familiarity with the equations of motion, including v = v0 + at.
  • Knowledge of trigonometric functions, particularly sine and cosine.
  • Ability to apply the Pythagorean theorem in physics contexts.
NEXT STEPS
  • Study the derivation and application of the equations of motion in projectile problems.
  • Learn how to analyze projectile motion using vector components.
  • Explore the effects of varying launch angles on projectile trajectories.
  • Investigate the impact of air resistance on projectile motion calculations.
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Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for examples of common student misconceptions in these topics.

  • #31
akshajkadaveru said:
HOLY MOLY...in all honesty.. am i over complicating this ?
Quite possibly.
 
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  • #32
so for D... i already have the x y values and the angle right? so now i just have to plot them or something >
 
  • #33
akshajkadaveru said:
so for D... i already have the x y values and the angle right? so now i just have to plot them or something >
Plot them if you wish, but to answer the question you should use algebra and calculation.
If the horizontal and vertical velocities are u and v respectively, what is the overall speed, and what is the angle of it?
 
  • #34
i know how to find the angle.. but how do i find the overall speed... is that by the pythagorean theorem and i plot that based on what i got for the V0x and v0y?
 
  • #35
akshajkadaveru said:
overall speed... is that by the pythagorean theorem
Yes.
 
  • #36
2^2+3.46^2=15.97
sqrt(15.97)
3.99 m/s
tan^-1(3.46/2)= 59.97 degrees
3.99m/s @ 59.97 degrees'
 
  • #37
akshajkadaveru said:
2^2+3.46^2=15.97
sqrt(15.97)
3.99 m/s
tan^-1(3.46/2)= 59.97 degrees
3.99m/s @ 59.97 degrees'
You've used the initial vertical velocity. As a result, you've recalculated the initial angle and overall speed.
Use the vertical velocity when it hits the target.
 
  • #38
ARGHGHGHH! so what i JUST did was wrong but what will my X and Y value things be then ?
 
  • #39
akshajkadaveru said:
ARGHGHGHH! so what i JUST did was wrong but what will my X and Y value things be then ?
You posted the correct final vertical velocity in post #27. Just use that instead of 3.46 in your calculation in post #36.
 
  • #40
i didn't post anything on 27
 
  • #41
akshajkadaveru said:
i didn't post anything on 27
I meant 28.
 
  • #42
so is it... 1^2+1.44^2=whatever
sqrt(whatever)= ______ @ tan^-1(1.44/1)?

will that be my answer?
 
  • #43
akshajkadaveru said:
so is it... 1^2+1.44^2=whatever
sqrt(whatever)= ______ @ tan^-1(1.44/1)?

will that be my answer?
Where are you getting 1m/s from for the x direction?
 
  • #44
oh... i thought that't just the length of how far the thing went... how do i get that then >?
 
  • #45
akshajkadaveru said:
oh... i thought that't just the length of how far the thing went... how do i get that then >?
No, you can't add a velocity to a length, or a velocity-squared to a length-squared. It makes no sense.
To apply the arctan formula and Pythagoras, you need two things of the same type at right angles - both lengths, or both velocities, or both accelerations...

What is the initial horizontal velocity? What is the horizontal acceleration? So what is the final horizontal velocity?
 

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