Kio asked in Science & MathematicsPhysics · 7 years ago

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

  • 7 years ago
    Favorite Answer



    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


    Find the total time the projectile is in the air using

    h = v×t + a×t²/2


    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 = ?


    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)


    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


    r = horizontal range = ?

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

    t = time = -2×Vo×sin(θ) / g


    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.

    • Login to reply the answers
  • 7 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

    • Login to reply the answers
  • 7 years ago

    of corse

    • Login to reply the answers
Still have questions? Get your answers by asking now.