Here is a physics question I've been stuck on?

A cannon with a muzzle velocity of 500. meters per second fires a cannonball at an angle of 30.° above the horizontal. What is the vertical component of the cannonball's velocity as it leaves the cannon?

3 Answers

  • 2 months ago
    Favorite Answer

    500sin(30º) = 250m/s

    If you want the understand why, try the video in the link.

  • 2 months ago

    v=500 sin(30)=500*.5=   250m/s           .

  • 2 months ago

    The components are V sin(angle) and V cos(angle). I can't automatically remember which is horizontal and which is vertical, so I have to think about it.  If the angle was zero, then the components would be 0 vertical, and V horizontal. So then, which one of these is equal to zero: V sin(0) and V cos(0)?

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