• Support PF! Buy your school textbooks, materials and every day products Here!

Projection angle using max. height w/ proportions only

  • #1
Figures that the only problem I have trouble with is the one the book considers to be "easy":

Homework Statement


The speed of a projectile when it reaches its maximum height is one half its speed when it is at half its maximum height. What is the initial projection angle of the projectile?


Homework Equations


h= (vi^2 sin^2 theta)/2g


The Attempt at a Solution



Complete brain fart--I don't even know where to start, beyond trying to divide h/2, which would be (vi^2 sin^2 theta)/4g, correct? That's about as far as I got. I just keep miring myself the further into it I go.
 

Answers and Replies

  • #2
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,812
6
Speed of course is the magnitude of velocity. At its maximum height a projectile only has one component of velocity. Is that enough to get you started?
 
  • #3
461
0
Try breaking velocity into x and y components @ 1/2 max height and then finding speed from there the "sqrt of the sum of the squares".
 

Related Threads for: Projection angle using max. height w/ proportions only

Replies
2
Views
12K
Replies
9
Views
3K
Replies
3
Views
11K
Replies
2
Views
25K
  • Last Post
Replies
1
Views
3K
Replies
3
Views
594
Top