Projectile Motion  Rock Thrown Horizontally Off a Cliffby Nickg140143 Tags: 2 dimension, motion, projectile, rock 

#1
Sep1611, 09:34 PM

P: 30

1. The problem statement, all variables and given/known data
A student stands on the edge of a cliff and throws a stone horizontally over the edge with a speed "v_{1}". The cliff is "h" meters high. Given [h,v_{1}], Determine: a. The time to hit the ground b. The horizontal distance traveled c. The magnitude and direction of the stone's velocity just before hitting the ground 2. Relevant equations x direction [tex] V_x=V_{0x} [/tex] [tex] x=V_0\cos{\theta} [/tex] y direction [tex] V_y=V_{0y}gt [/tex] [tex] y=V_{0y}t\frac{1}{2}gt^2 [/tex] [tex] V_y^2=V_{0y}^22gy [/tex] 3. The attempt at a solution If I set up my coordinate system so that 0 is ground level, and h is cliff level (where the stone was thrown from), and the distance to the landing point of the stone when it hits the ground is, say "r" meters from 0 in the x direction. I've shown what work I have in the attached images, but to be perfectly honest, I'm having trouble with the entire basic strategy to approaching this problem. When I try to solve for x distance r, I need time. So when I try to solve for time, that becomes reliant on r. I can't even think of anything else I can find. I feel I'm either not utilizing the possible angles within this problem, or I've completely missed some fundamental idea in regards to projectile motion. A good shove in the right direction on this kind of problem would be VERY much appreciated. 



#2
Sep1611, 10:02 PM

P: 3,628

Here's a simple approach:
Firstly make a table of Horizontal motion and vertical motion, like according to your question, HHorizontal motion VVertical motion uinitial velocity sdisplacement hHeight of the cliff x and t are assumed since we have to find them out. Ok, first lets see in which column, assumed variable are less. Its obvious that in V column, there is only one variable "t". So can you apply equation of motion here? 



#3
Sep1611, 10:19 PM

P: 30

Something tells me I've once again forgotten a rather important fact about the equation
[tex]y=V_{0y}t+\frac{at^2}{2}[/tex] Is it correctly written as this by chance [tex](yy_0)=V_{0y}t+\frac{at^2}{2}[/tex] 



#4
Sep1611, 10:21 PM

P: 3,628

Projectile Motion  Rock Thrown Horizontally Off a CliffNow just plug in the values and you will get the time. :) 



#5
Sep1611, 10:37 PM

P: 30

How about this then:
[tex](yy_0)=V_{0y}t+\frac{at^2}{2}[/tex] [tex](0h)=0\frac{gt^2}{2}[/tex] [tex]2h=gt^2[/tex] [tex]\frac{2h}{g}=t^2[/tex] [tex]\sqrt{\frac{2h}{g}}=t[/tex] 



#6
Sep1611, 10:40 PM

P: 3,628

That's right.
Now just see that we have t same for both H and V motion. So now we are left with x in H motion. Try applying the suitable equation of motion in the H column. 



#7
Sep1611, 10:50 PM

P: 30

Here is my attempt for Horizontal distance covered:
[tex]x=(V_0\cos{\theta})t[/tex] [tex]x=V_1t[/tex] [tex]x=\frac{V_1\sqrt{2h}}{g}[/tex] hows that look? Now in order to get the direction and magnitude of the end velocity of the rock, I need both the V_{x} and V_{y} right? 



#8
Sep1611, 10:53 PM

P: 3,628

Now you have solved the first two parts. About the third part, you can go like this: Find the final velocity in Horizontal and Vertical direction, then find their resultant. :) 



#9
Sep1611, 11:11 PM

P: 30

Alright, so far so good, lets see if I got this right:
To find the Velocity in the y direction near point of impact [tex]V_y=V_{0y}^2+2a(yy_0)[/tex] substitute in my knowns: [tex]V_y=(0)g(\frac{\sqrt{2h}}{g})[/tex] [tex]V_y^2=(0)2g(0h)[/tex] [tex]V_y^2=2g(h)[/tex] [tex]V_y^2=2gh[/tex] [tex]V_y = \sqrt{2gh}[/tex] As for the resultant, are you referring to vector addition? If so, I think I got [tex]<V_1,0,0>+<0,\sqrt{2gh},0>\ = \ <V_1,\sqrt{2gh},0>[/tex] Hows all of this look? But what if the question asked for the angle, would I need to use the Tangent Function and take the Inverse Tangent to find Theta? 



#10
Sep1611, 11:19 PM

P: 3,628

What i expected was that you would have used the first equation of motion. Yep, i am referring to vector addition. I think you did it by coordinates. I have never used that way. We have horizontal and vertical final velocities. These both are perpendicular to each other. So [tex]V=\sqrt{V_x^2+V_y^2}[/tex] V is the resultant velocity. V_{x}=v_{1} V_{y}=[itex]\sqrt{2gh}[/itex] Just plug in the values and your are done. If you want to find the angle, you need to first specify the reference, Horizontal or vertical. :) 



#11
Sep1611, 11:23 PM

P: 30

Well, it seems I got a grasp of the problem now, I think the only thing that messed me up in the beginning was the fact that I was applying one of the formulas wrong.
Thanks for all the help, its much appreciated 



#12
Sep1611, 11:35 PM

P: 3,628

And remember the table approach. Whenever you deal with projectiles, the best way is to make the table I forgot to welcome you. Welcome to the board. Hope you enjoy your stay here. Before wandering on the board, have a look at the rules. They may help you. 



#13
Sep1711, 12:08 AM

P: 30

Thanks for the advice
Thanks again 


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