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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Answered: Projectile motion in 2D A projectile is…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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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

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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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