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Homework Help: Projectile Motion Graph Help

  1. Jan 30, 2006 #1
    Hi guys, I have a Homework problem that is related to Projectile Motion. It is said a projectile is fired form a surface(level ground) at an angle A, above the horizonal. so here is the question, there is another angle B, which reachs the same Max high with angel A, how can I prove that

    The graph showed the path of angle A is a paraola and has a range R, but the path of angle B is only a stright line to the Max High which is H.

    I know that Time or the vertical V is going to be the key since they used the same time to reach the same high, but I just couldn't get it right, with all the subs. so would u please give me a hint for that? thanks much!
  2. jcsd
  3. Jan 30, 2006 #2


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    What you are talking about, angle B, is the angle of elevation of the point on the trajectory of the projectile where it reaches its maximum height. It can be found from the ratio of [itex]\frac{2H}{R}[/itex] which gives you the tangent of angle B.
    Last edited: Jan 30, 2006
  4. Jan 30, 2006 #3
    Well I didn't follow you. Two projections will have same value of maximum height attained only if they have the same vertical component.
  5. Jan 30, 2006 #4


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    I agree with vaishakh, I can't see how this would work
  6. Jan 30, 2006 #5
    I got it. thanks andrevh. Maybe I didn't express the question right. there was a graph, it would be much easier to understand. but thanks for your help anyway (vaishakh, hootenanny)
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