Projectile motion Problem(Is My Ans Correct?)

[Plutonium88]

The question: An object has a Velocity V1 and an angle A (Alpha) initially. Later in its path, this same object has an Angle B (Beta)

find the time elapsed from point A to point B


Okay so i have an answer.. i just need some one to tell me if what I'm doing is correct./answer is correct.

http://postimage.org/image/xsei89eg1/


here is an image of the problem.


Okay so..

V1x = V1cosA
V1y = V1sinA

Let V2 be the variable for the second velocity.

Since horizontal component of velocity is constant through projectile motion..

V1x=V2x

V2x = v1cosA

Through trigonemtry we can resolve the Y component for V2, which i need for further calculation

TanB = V2y/V2x

V2y= V2xTanB

Now i have all i need and i can use uniform accelerated formula, to solve for The time...

Vf=Vi + gt

V2y = V1y + gt

(v2y - v1y)/g = t

Answer: t = (v1cosATanB - V1sinA)/g


or simplifieed too... V1(cosA*TanB - SinA)/g

(m/s)/(m/s^2)

= s (seconds)

Units match.

 

[Delphi51]

Got the same thing with opposite sign - I took g to be 9.81; in your solution g = -9.81.

 

[Plutonium88]

lol i always make taht stupid mistake.. xD.

so t= (v1cosAtanB - v1sina) /-g

t = (v1sina - v1cosAtanB )/ g

 

[Plutonium88]

Got the same thing with opposite sign - I took g to be 9.81; in your solution g = -9.81.

d

so using your frame of reference for the vertical component as g = 9.81 and X portion being positive in the direction of motion
v2x=v1x

TanB = -V2y/v2x

-V2y = TanB*V1cosA
v2y = -TanB*v1cosA

v2y = v1y + gt

-TanB*v1CosA = -V1SinA + Gt
T = (V1SinA - V1CosA*TanB)/g

T= V1(SinA-CosA*TanB)/g


I'm just curious for what i did in the post above, am i 'allowed' to do that..

cause i thoguht -a/b is the same thing as a/-b

? Also i prefer using your way especially when using purely variables making;
Gravity = Positive Direction

because i don't have to account for the sign, therefore i won't lose track of it.