Where on the hill does the projectile land?

[TarPaul91]

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.

 

[Orodruin]

How would you describe the y-position of the projectile?

 

[TarPaul91]

How would you describe the y-position of the projectile?

Would it be 79(sin 60) t or am I getting the concepts confused?

 

[Orodruin]

That would be correct if there was no gravity.

 

[TarPaul91]

That would be correct if there was no gravity.

Oh, thanks. So then how do I factor for gravity?

 

[Orodruin]

What are your own thoughts regarding including gravity?

 

[TarPaul91]

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?

 

[Orodruin]

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 ...

 

[TarPaul91]

Oh, the g should be negative, yes?

 

[Orodruin]

Oh, the g should be negative, yes?

You tell me. What would it mean if it was positive/negative?