Supposedly easy physics problem =( [projectile]

[akshajkadaveru]

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'

 

[haruspex]

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.

 

[akshajkadaveru]

ARGHGHGHH! so what i JUST did was wrong but what will my X and Y value things be then ?

 

[haruspex]

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.

 

[akshajkadaveru]

i didn't post anything on 27

 

[haruspex]

i didn't post anything on 27

I meant 28.

 

[akshajkadaveru]

so is it... 1^2+1.44^2=whatever
sqrt(whatever)= ______ @ tan^-1(1.44/1)?

will that be my answer?

 

[haruspex]

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?

 

[akshajkadaveru]

oh... i thought that't just the length of how far the thing went... how do i get that then >?

 

[haruspex]

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?