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# Projectile motion question?

Ok, so I understand for projectile motion questions, if given the initial velocity, we must break that into it's x and y components.

1) To completely solve the problem, how do we put them back together?

2) How do we find the range after finding those, but no time t is available.

### 3 Answers

- JánošíkLv 77 years agoFavorite Answer
1)

say

initial velocity = Vo

launch angle = θ

horizontal component = Vh

vertical component = Vv

When knowing any 2 of Vo, θ, Vh, Vv, you can find the other 2:

knowing Vo and θ:

Vh = Vo×cos(θ)

Vv = Vo×sin(θ)

knowing Vo and Vh:

θ = arccos(Vh/Vo)

Vv = √( Vo² - Vh² )

knowing Vo and Vv:

θ = arcsin(Vv/Vo)

Vh = √( Vo² - Vv² )

knowing θ and Vh:

Vo = Vh / cos(θ)

Vv = Vo×tan(θ)

knowing θ and Vv:

Vo = Vv / sin(θ)

Vh = Vv / tan(θ)

knowing Vh and Vv:

Vo = √( Vv² + Vh² )

θ = arctan(Vv/Vh)

You can easily find these formula's yourself by drawing a right triangle where

Vo = hypotenuse

Vv = opposite side of θ

Vh = adjacent side of θ

2)

Find the total time the projectile is in the air using

h = v×t + a×t²/2

where

h = height = 0 m (you need the time when it is back at the ground)

v = initial vertical velocity = Vv = Vo×sin(θ)

a = acceleration by gravity = g (=-9.8 m/s²)

t = time = ?

so

0 = Vo×sin(θ)×t + g×t²/2

0 = (Vo×sin(θ) + g×t/2)×t

t = 0 (at time t=0 the projectile is still at height 0 m)

or

Vo×sin(θ) + g×t/2 = 0

t = -2×Vo×sin(θ) / g (at that time the projectile is back at height 0 m)

Now look at the horizontal range:

r = v × t

where

r = horizontal range = ?

v = horizontal velocity = Vh = Vo×cos(θ)

t = time = -2×Vo×sin(θ) / g

so

r = (Vo×cos(θ)) × (-2×Vo×sin(θ) / g)

r = -(Vo)² × sin(2θ) / g

This is the horizontal range as a function of Vo and θ and the constant g (=-9.8 m/s²)

Note the minus sign in front of the formula for r. It will be canceled out by the minus sign of the value for g.

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- Shaun DizzleLv 77 years ago
To put them back together they are simply put together via the pythagorean theorem:

√((vx)^2 + (vy)^2) = v

To find the range when no time is given we use the equation:

R = (vo)(sin2ø) / g

The proof for it is rather simple and you should not have a problem looking it up and understanding it. If you still draw a blank shoot me an email and ill try and explain it

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