My question is that I don’t understand what happened in the second page from where it says we obtain:I get how we got X but I don’t get how we get alpha in the first page and then use it in the second page till we get the final formula if you can please show me the steps of finding the final formula Projectile motion in 2DA projectile is launched from the edge ofa cliff, at a height yo above the underlyingplateau. The initial velocity vector V=V₂ î+ Vyomakes an angle = 45° with respect to thehorizontal, as shown in the figure. Theprojectile lands at a distance from thebottom of the cliff. Find the vebuity vyo asa function of X=xyo and Yogmakes this possible.that9yaLO2We obtain:1-91AGt=2₂X = (1-11-2a) - 3A few algebraic steps:Xa-1 =√1+2aX2²-2X2 + 1 = 1+22Xα = 2(x+1)a= 2 (2x²) = 3²j²X+1Newton's II law with constant force (10)Equation of motion: md=-mg 5₁ )= V₁dtmdv₁ = 0 → √x = √₂0, x(+) = √₂stdtdvymd=-mg → vy=vy-gt, y=y+t-1dt.The projectile touches ground at time t= tosuch that≈ 0.3s3Therefore:,20-%-%-²5- ² - 1 3 - 34²-(1-/14+ 26²)t₁ ==g² g(-sign rejected as it gives to <0).V74= -2²0 (1-/14 364)XG=1+Now use =V₂ (for 9=45") and introduceX = 24₂Yoand a gV²X = 2Yog = 2400 m²/s²Numbers:Input parameters are y. and X= 7₂/9₁,while g= 3.8 m/s². To make things simple,chooseThenVo = x 2400 = x= 16×10² m²Vyo = 40 mThe three component of this vectorequation are45

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Projectile motion in 2D A projectile is launched from the edge of a cliff, at a height yo above the underlying plateau. The initial velocity vector V=V₂₁ ²+ VyoĴ makes an angle = 45° with respect to the horizontal, as shown in the figure. The projectile lands at a distance from the bottom of the cliff. Find the vebuity vyo as a function of X=xel yo and yog makes this possible. that 9. لم YA L We obtain: 1-95 AG X = = = (1+11+20) A few algebraic steps: Xa-1 =√1+2a Vy = / Yog X2²-2X2+1=X+22 Xα = 2(x+1) d = 2 = 275 yo 34 Newton's II law with constant force (10) Equation of motion: md=-mg )= V dt m → Vx=V₂0, X(t) = √₂+ dvy m мон o-ng - Yo Yongt, утунцаћи де dt. dv₁ = 0 dt The projectile touches ground at time t= ta such that 0=yo+vyotg Vyeo ± √ V 24 Vy g² g = (-sign rejected as it gives to <0). Therefore: XG = V Yoyo (1+1/1+ 2002) g Now use =V₂ (for 9=45°) and introduce and d = X = 26 yo Numbers: while Input parameters are y. and X= 7₂/9₁, g=9.8 m/s². To make things simple, choose X = 2 Yog Then = 2400 m/s² vz Vyo = 40 The three component of this vector equation as 5

Projectile motion in 2D A projectile is launched from the edge of a cliff, at a height yo above the underlying plateau. The initial velocity vector V=V₂₁ ²+ VyoĴ makes an angle = 45° with respect to the horizontal, as shown in the figure. The projectile lands at a distance from the bottom of the cliff. Find the vebuity vyo as a function of X=xel yo and yog makes this possible. that 9. لم YA L We obtain: 1-95 AG X = = = (1+11+20) A few algebraic steps: Xa-1 =√1+2a Vy = / Yog X2²-2X2+1=X+22 Xα = 2(x+1) d = 2 = 275 yo 34 Newton's II law with constant force (10) Equation of motion: md=-mg )= V dt m → Vx=V₂0, X(t) = √₂+ dvy m мон o-ng - Yo Yongt, утунцаћи де dt. dv₁ = 0 dt The projectile touches ground at time t= ta such that 0=yo+vyotg Vyeo ± √ V 24 Vy g² g = (-sign rejected as it gives to <0). Therefore: XG = V Yoyo (1+1/1+ 2002) g Now use =V₂ (for 9=45°) and introduce and d = X = 26 yo Numbers: while Input parameters are y. and X= 7₂/9₁, g=9.8 m/s². To make things simple, choose X = 2 Yog Then = 2400 m/s² vz Vyo = 40 The three component of this vector equation as 5

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter2: Vectors

Section: Chapter Questions

Problem 61P: In the control tower at a regional airport, an air traffic controller monitors two aircraft as their...

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