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Angle at which distance is longest?

  1. Nov 2, 2008 #1
    I have a question about projectory motion.

    I know that if ball is thrown by a person with a initial velocity v(only x and ycomponents), a maximum distance d (distance of x component) can be achived at angle of 45 degree (an angle between the initial velocity and x component of the velocity). But this is only true when the initial position of y component is the same as the terminal position of y conpoment. Here's the my question: how do you guys find the angle at which when the height of a landing position is not the same as the height of the initial position (i.e. Initial y is not equal to terminal y)? Let's say the ball lands on a tower which has a height h with minimum velocity. What's the formula to find the angle?
  2. jcsd
  3. Nov 2, 2008 #2
    You mean, what is the angle for the ball to land on the tower? Or to reach maximum range?
    Or you have a series of towers and you want to reach the farthest tower possible?
  4. Nov 7, 2008 #3
    Does the attachment answer your question?

    Attached Files:

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