Projectile Fired Against A Slope

  • #1

Homework Statement



A projectile is launched at an angle 45 degrees above the horizon with a velocity of 30 m/s. It is launched at the base of a constant slope, which has an angle of 30 degrees.
a- How much time did it take for the projectile to hit the slope?
b- Where did the projectile hit the slope? Use a vector.
c- Was the projectile moving up or down when it hit?


Homework Equations



t = 2 V sin angle /g

x= V^2 sin 2angle / g

The Attempt at a Solution




a- (2(30 m/s) sin 15)/ 9.8 m/s^2
t = 1.58 s

b- (30 m/s (cos 15) ) i + (30 m/s (sin 15)) j

I am not sure I should be using 15 as the angle for part b. I tried with 30 and 45, and 30 seems like the right coordinates considering the picture I drew.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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Homework Statement



A projectile is launched at an angle 45 degrees above the horizon with a velocity of 30 m/s. It is launched at the base of a constant slope, which has an angle of 30 degrees.
a- How much time did it take for the projectile to hit the slope?
b- Where did the projectile hit the slope? Use a vector.
c- Was the projectile moving up or down when it hit?


Homework Equations



t = 2 V sin angle /g

x= V^2 sin 2angle / g

The Attempt at a Solution




a- (2(30 m/s) sin 15)/ 9.8 m/s^2
t = 1.58 s

b- (30 m/s (cos 15) ) i + (30 m/s (sin 15)) j

I am not sure I should be using 15 as the angle for part b. I tried with 30 and 45, and 30 seems like the right coordinates considering the picture I drew.

You shouldn't use the angle 15 at all.

You should resolve your Velocity into x,y.

Then write your y height equation.

You know the height of the slope from its angle and the x distance traveled. So for each m in the x direction gives you a half meter rise in slope (sin30), you can translate x-velocity into y height by:

y = Vx*t*sin30 = V*cos45*sin30*t

But y also = vy*t - 1/2*a*t2 = V*sin45*t - 1/2*a*t2

With sin30 = 1/2 and sin45 = cos45 = 1/(2)1/2, solving is pretty straight forward.
 
  • #3
I'm sorry, I really don't understand what you mean. Are you creating these equations to solve for t? Doing that, I get t = 2.17 s. Does that mean the time I found using the range equation is wrong? I also don't understand how to put the directions in vectors.
 
  • #4
LowlyPion
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I'm sorry, I really don't understand what you mean. Are you creating these equations to solve for t? Doing that, I get t = 2.17 s. Does that mean the time I found using the range equation is wrong? I also don't understand how to put the directions in vectors.

That's what I get for time (actually 2.1642, but close enough).

Part b is asking for the distance up the slope. It is a displacement vector pointing up the slope. Knowing the x distance component at t=2.17 then the magnitude can be figured as to how it translates up the slope.

Part c is asking whether it hit before max height or after. Calculate time to max height. Is 2.17 greater or less?
 
  • #5
But how do I know the x-distance component? Why was it wrong to use the formula I had before?
 
  • #6
LowlyPion
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But how do I know the x-distance component? Why was it wrong to use the formula I had before?

The x distance is V*cos45*t isn't it? That is how far along the x axis it gets at impact. But it hit on the slope at a height Y determined by using Cos30 degrees = x distance, what is the distance up the hypotenuse. That's your distance up the slope.

I'm sorry but I didn't really look at your first formulas after I saw functions of 15 degrees, since I thought this was the easier way to write and solve the problem.
 
  • #7
OK. I'm sorry this is taking me so long to understand. I see why the x-component is cos 45, but I still don't know how to determine the y height from cos 30. If I am doing this correctly, I get 21.21 m for the x displacement. The using a triangle, I find 12.24 for y. Is that what you meant?
 
  • #8
LowlyPion
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OK. I'm sorry this is taking me so long to understand. I see why the x-component is cos 45, but I still don't know how to determine the y height from cos 30. If I am doing this correctly, I get 21.21 m for the x displacement. The using a triangle, I find 12.24 for y. Is that what you meant?

And then the distance along the hypotenuse is the vector. Which is twice 12.24 right?

But those would be the x,y components of the vector.

The cos 30 comes from the angle of the slope. The adjacent side is the Cos 30 = the x distance traveled.
 
Last edited:
  • #9
OK, I think I understand it now. Thanks so much for your time!
 
  • #10
LowlyPion
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OK. I'm sorry this is taking me so long to understand. I see why the x-component is cos 45, but I still don't know how to determine the y height from cos 30. If I am doing this correctly, I get 21.21 m for the x displacement. The using a triangle, I find 12.24 for y. Is that what you meant?

Wait a minute. 21.21 is only the x-distance of the velocity in 1 sec. This needs to be multiplied by the time to determine x distance at impact. And then the hypotenuse by dividing by Cos30. Sorry I missed that.
 
  • #11
That's OK. That answer (45.8 m for x ) is what I got using the range equation as I was in the beginning. Is there really any reason I couldn't have done it that way?
 
  • #12
LowlyPion
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That's OK. That answer (45.8 m for x ) is what I got using the range equation as I was in the beginning. Is there really any reason I couldn't have done it that way?

I don't know that formula. Did you derive that or was it in the book? (I'm not feeling up to deriving it right now myself.)

I just went with what I saw and that method should get you to the right result.

Sorry if I misled you in any way.
 
  • #13
Don't worry about it. I found that formula in the book in the section on "range equations" I figured I could just act as if the projectile were being fired at a 15 degree angle and ignore the rest.
 

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