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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
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
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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 2D A projectile is launched from the edge of a cliff, at a height yo above the underlying plateau. The initial velocity vector V=V₂ î+ Vyo 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=xyo and Yog makes this possible. that 9 ya L O 2 We obtain: 1-91 AG t=2₂ X = (1-11-2a) - 3 A few algebraic steps: Xa-1 =√1+2a X2²-2X2 + 1 = 1+22 Xα = 2(x+1) a= 2 (2x²) = 3² j² X+1 Newton's II law with constant force (10) Equation of motion: md=-mg 5₁ )= V₁ dt m dv₁ = 0 → √x = √₂0, x(+) = √₂st dt dvy md=-mg → vy=vy-gt, y=y+t-1 dt. The projectile touches ground at time t= to such that ≈ 0.3s 3 Therefore: ,2 0-%-%-² 5- ² - 1 3 - 34²-(1-/14+ 26²) t₁ = = g² g (-sign rejected as it gives to <0). V 74= -2²0 (1-/14 364) XG= 1+ Now use =V₂ (for 9=45") and introduce X = 24₂ Yo and a g V² X = 2 Yog = 2400 m²/s² Numbers: Input parameters are y. and X= 7₂/9₁, while g= 3.8 m/s². To make things simple, choose Then Vo = x 2400 = x= 16×10² m² Vyo = 40 m The three component of this vector equation are 4 5
Transcribed Image Text: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=V₂ î+ Vyo 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=xyo and Yog makes this possible. that 9 ya L O 2 We obtain: 1-91 AG t=2₂ X = (1-11-2a) - 3 A few algebraic steps: Xa-1 =√1+2a X2²-2X2 + 1 = 1+22 Xα = 2(x+1) a= 2 (2x²) = 3² j² X+1 Newton's II law with constant force (10) Equation of motion: md=-mg 5₁ )= V₁ dt m dv₁ = 0 → √x = √₂0, x(+) = √₂st dt dvy md=-mg → vy=vy-gt, y=y+t-1 dt. The projectile touches ground at time t= to such that ≈ 0.3s 3 Therefore: ,2 0-%-%-² 5- ² - 1 3 - 34²-(1-/14+ 26²) t₁ = = g² g (-sign rejected as it gives to <0). V 74= -2²0 (1-/14 364) XG= 1+ Now use =V₂ (for 9=45") and introduce X = 24₂ Yo and a g V² X = 2 Yog = 2400 m²/s² Numbers: Input parameters are y. and X= 7₂/9₁, while g= 3.8 m/s². To make things simple, choose Then Vo = x 2400 = x= 16×10² m² Vyo = 40 m The three component of this vector equation are 4 5
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my question was also how did we get alpha to begin with so that we can substitute it

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