Projectile Motion (find theta)

  • #1
bengavinb20
2
0

Homework Statement



The speed of a projectile when it reaches its maximum height is 0.31 times 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 θ) / 2g

R=(Vi^2 * sin2θ) / g

The Attempt at a Solution



I realize that i need to somehow come up with a ratio and set the two equal to each other. However, i'm having difficulty with the initial step. Maybe I could get a hint that could set me on the right track?
 

Answers and Replies

  • #2
yands
35
2
Decompose the velocity vector.
At the top which component of velocity is zero?
Use this idea to find the ratio. As you mentioned
 
  • #3
bengavinb20
2
0
So after decomposing the velocity vector I got:

Vi = -yf-at2 /t

Plugging that in to the equation for h...

h= ((-yf-at2)2*(sin2θ)) / 2g

and then

0.5h = 0.31((-yf-at2)2*(sin2θ)) / 2g

Am I on the right track?
 
  • #4
yands
35
2
You see at the top point V is just Vx so can you relate this to the given information.
 
  • #5
yands
35
2
I'm sorry for late response I got lectures :)
So you still solving the problem?
 

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