High School Physics: Projectiles at Angles

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Geminiphysics
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Homework Statement



"A ball rolls off an incline plane. The speed of the ball is 22m/s and the plane is inclined at 32 degrees to the horizontal.If the incline is 9.0m above the ground, how far from the base of the incline will the ball hit the floor?" (The rounded answer is supposed to be 11m)
There is also a diagram that goes with this question, which I have attached a picture of.

Homework Equations



Cosθ, sinθ, or tanθ?

The Attempt at a Solution



The rounded answer is supposed to be 11m.
The closest answer I got to this was 10.6126...m which I accidentally got by 9/(cos32). After thinking it over, I realized that I was calculating the length of the ball's vertical path (according to my understanding of the diagram. I drew what I believed to be the correct diagram and attached it, but I don't think it is right anymore.) I'm mostly confused about the placement of the 32° because, from my understanding, I think it has to be placed at another part of the diagram, other than on the roof.
Thank you in advance!
 

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Your supplied diagram is upside-down.
 
those two angles can't be 32 can they with a 90 in the center? if one is 32 the other has to be 58
 
462chevelle said:
those two angles can't be 32 can they with a 90 in the center? if one is 32 the other has to be 58

They're in two separate triangles, it's fine.

EDIT:

lol after holding my head upside down I see your point now. I'm thinking that his drawn-in 32 degree angle should be 58 degrees, like you said.
 
ya. the hyp. is a 180 angle and has 3 angles that must =180
 
Thank you for your suggestions. I will try them and see if I can figure it out!
 
This time I tried putting the 32° in the bottom right hand corner. (I attached the picture of my new diagram)
Then, I solved for the Vertical velocity of the ball by using sinθ= Vv(vertical velocity)/Vr(resultant velocity).
I got Vv= (sin 32)(22m/s). With this I got the answer of 11.6582...m/s for the vertical velocity.
Then I figured out the horizontal velocity (Vh) by using cosθ=Vh/Vr
With the substitutions and some manipulation it became Vh= (cos32°)(22m/s) which equals 18.657...m/s.
I was thinking of using the formula d=vt+1/2at^2 to find the time (to multiply with the horizontal velocity to get the total horizontal distance) but then I realized that I would have to know the quadratic formula to do so... Which I don't...
So I was wondering if there's any other formula I could use to solve for time and if what I have done so far is correct. Thanks again!
 

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¡sʞuɐɥʇ ˙ɹǝısɐǝ ɥɔnɯ sʇןnsǝɹ ɹnoʎ buıʇǝɹdɹǝʇuı puɐ buıpɐǝɹ sǝʞɐɯ ʇı ؛ʇı ǝʇɐıɔǝɹddɐ ʎןןɐǝɹ pןnoʍ ǝʍ
 
Sorry about that, I'll attach another picture that's hopefully not upside-down.
 

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Geminiphysics said:
Sorry about that, I'll attach another picture that's hopefully not upside-down.

Better :approve:

So you know the initial velocity in the y-direction, the height in the y-direction (9m), the acceleration (obviously 9.8 m/s2 due to gravity), and what you really want to find is the time that the ball is in the air, because knowing the time and velocity of the ball (in the x-direction) allows for a simple computation to find its distance traveled.

Look for kinematics equations that will allow for you to find the time that the ball is in the air using the information that you do have (edit: I won't supply you with them because a majority of the difficulty in these problems is knowing how to manipulate the equations you have, and to be able to recognize which equations are useful in which situations. Good luck! :smile:)
 
Thank you everyone! I finally figured out the answer! I appreciate all of your time and help!
 
awesome.