- 42,843
- 10,509
Quite possibly.akshajkadaveru said:HOLY MOLY...in all honesty.. am i over complicating this ?
The discussion revolves around a projectile motion problem where a projectile is launched with an initial velocity at a specific angle and is required to hit a target at a given horizontal distance. Participants are exploring various aspects of the projectile's motion, including its vertical and horizontal components, time of flight, and final velocities.
Some participants have provided guidance on the equations of motion and have prompted others to clarify their understanding of acceleration and velocity. There appears to be a mix of correct and incorrect assumptions, with ongoing attempts to reconcile different interpretations of the problem and calculations. The discussion is active, with participants seeking clarification and exploring different approaches.
Participants are navigating through potential confusion regarding the time of flight and the relationship between horizontal and vertical motion. There are indications of differing interpretations of the problem setup and the values to be used in calculations, which may affect the overall understanding of the projectile's motion.
Quite possibly.akshajkadaveru said:HOLY MOLY...in all honesty.. am i over complicating this ?
Plot them if you wish, but to answer the question you should use algebra and calculation.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 >
Yes.akshajkadaveru said:overall speed... is that by the pythagorean theorem
You've used the initial vertical velocity. As a result, you've recalculated the initial angle and overall speed.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 posted the correct final vertical velocity in post #27. Just use that instead of 3.46 in your calculation in post #36.akshajkadaveru said:ARGHGHGHH! so what i JUST did was wrong but what will my X and Y value things be then ?
I meant 28.akshajkadaveru said:i didn't post anything on 27
Where are you getting 1m/s from for the x direction?akshajkadaveru said:so is it... 1^2+1.44^2=whatever
sqrt(whatever)= ______ @ tan^-1(1.44/1)?
will that be my answer?
No, you can't add a velocity to a length, or a velocity-squared to a length-squared. It makes no sense.akshajkadaveru said:oh... i thought that't just the length of how far the thing went... how do i get that then >?