Where on the hill does the projectile land?

  • #1
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Homework Statement


A projectile is shot at a hill, the base of which is 320 m away. The projectile is shot at 60° above the horizontal with an initial speed of 79 m/s. The hill can be approximated by a plane sloped at 19° to the horizontal. The equation of the straight line forming the hill is
y = x tan(19°) − 110.
Where on the hill does the projectile land? (Answer is supposed to be in the form x, y.)

Homework Equations


y = x tan(19°) − 110.

The Attempt at a Solution


Write expressions for the x and y coordinates of the projectile as a function of time.
x = 79 cos(60°) t
I wasn't sure where to go from here.
 

Answers and Replies

  • #2
Orodruin
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How would you describe the y-position of the projectile?
 
  • #3
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How would you describe the y-position of the projectile?
Would it be 79(sin 60) t or am I getting the concepts confused?
 
  • #4
Orodruin
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That would be correct if there was no gravity.
 
  • #5
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That would be correct if there was no gravity.
Oh, thanks. So then how do I factor for gravity?
 
  • #6
Orodruin
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What are your own thoughts regarding including gravity?
 
  • #7
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What are your own thoughts regarding including gravity?
Can I use the formula here d= (initial velocity)(time)+1/2(g)(t^2) or am I totally going in the wrong direction here?
 
  • #8
Orodruin
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Can I use the formula here d= (initial velocity)(time)+1/2(g)(t^2) or am I totally going in the wrong direction here?
You are going in the right direction, but something in your formula is not ...
 
  • #9
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Oh, the g should be negative, yes?
 
  • #10
Orodruin
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Oh, the g should be negative, yes?
You tell me. What would it mean if it was positive/negative?
 

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