Proving the Relationship Between Projectile Motion Angles Using Graphs

AI Thread Summary
The discussion revolves around proving the relationship between two projectile motion angles, A and B, which reach the same maximum height. The key point is that angle B, which has a straight-line path to the maximum height, can be expressed in terms of angle A using the equation tanB = (1/2)tanA. Participants note that both angles must have the same vertical component to achieve the same maximum height. The conversation highlights the importance of understanding the time and vertical velocity involved in projectile motion. Clarifications about the graph and trajectory paths were also mentioned to aid comprehension.
team31
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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
tanB=(1/2)tanA?

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!
 
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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 \frac{2H}{R} which gives you the tangent of angle B.
 
Last edited:
team31 said:
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
tanB=(1/2)tanA?
Well I didn't follow you. Two projections will have same value of maximum height attained only if they have the same vertical component.
 
I agree with vaishakh, I can't see how this would work
 
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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