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M1 projectiles, quick question

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    • Thread Starter

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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?

    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.
    • Thread Starter

    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'.
    • Thread Starter


    Well the horizontal is the horizontal line and your angle is below that

    Isn't projectile motion M2?

    (Original post by Sarabande)
    Isn't projectile motion M2?
    I guess it must be board dependant

    It is M1 for AQA

    (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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