- #1
Plutonium88
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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.
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.
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