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M1 projectiles, quick question

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    Part e) State the magnitude and direction of the velocity of the ball as it hits the surface.

    I've worked out the magnitude (20 ms-1) but I don't understand how to work out the direction. What does '50 degrees below the horizontal' mean?
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    If you have the magnitude of the horizontal and vertical components of velocity of the ball as it hits the surface, then you can form a triangle and use simple trig to find the angle theta below the horizontal.
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    Is this correct?

    Vertical velocity: y=20sin50 -9.8*3.13 = -15.4 ms-1
    Horizontal velocity x = 20cos50 = 12.9 ms-1

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    a = tan-1 (15.4/12.9) = 50degrees

    I'm still not sure what it means by 'below the horizontal'.
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    Anyone?
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    Well the horizontal is the horizontal line and your angle is below that
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    Isn't projectile motion M2?
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    (Original post by Sarabande)
    Isn't projectile motion M2?
    I guess it must be board dependant

    It is M1 for AQA
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    (Original post by TenOfThem)
    I guess it must be board dependant

    It is M1 for AQA
    Ah. For Edexcel it's the first kinematics topic of M2.

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Updated: April 16, 2012
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