Physics

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3. A projectile is launched at 10.0 m/s in a direction 45.0 degrees above the horizon. It hits the ground 10.0 m below its initial position. What is its speed just before hitting the ground? Ignore air resistance.

The answer is supposed to be 17.2 m/s.

  • Physics -

    I've been using:

    v(f)^2 = v(i)sin(45)^2 - 2g(y-y(i))

    Then I use:

    v = sqrt((v(f)^2 + v(i)^2)

    However, doing this gives me 18.6 m/s instead of 17.2 m/s. Any help would be greatly appreciated.

  • Physics -

    I've found that:

    v(f)^2 = v(i)^2 -2g(y - y(i))

    gives me 17.2 m/s. However, this way doesn't account for the angle, so I don't think it's the right way to solve this.

  • Physics -

    You've to consider the motion in vertical and horizontal directions separately.
    A)Vertical (y) dir.=>
    Vy^2 = Uy^2 + 2*(-g)*s
    = (10Sin45)^2 + 2*(-9.8)(-10)
    = 50 + 196 = 246
    So, Vy= 15.68 m/s
    (Vy is the vertical component of the projectile when it hits the ground)
    B) Horizontal (x)dir. =>
    Vx= Ux = 10Cos45 = 7.07 m/s

    Projectile's vel. just before it hits the gnd.=>
    V = sqrt(Vy^2 + Vx^2)
    = sqrt(246 + 50)
    = 17.2 m/s

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