Projectile Motion - Rock Thrown Horizontally Off a Cliff

  • #1

Homework Statement


A student stands on the edge of a cliff and throws a stone horizontally over the edge with a speed "v1". The cliff is "h" meters high. Given [h,v1], 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


Homework 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}^2-2gy
[/tex]

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.
 

Attachments

Answers and Replies

  • #2
3,812
92
Here's a simple approach:-

Firstly make a table of Horizontal motion and vertical motion, like according to your question,
2qmkthd.png


H-Horizontal motion
V-Vertical motion
u-initial velocity
s-displacement
h-Height 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? :smile:
 
  • #3
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 :grumpy:
[tex](y-y_0)=V_{0y}t+\frac{at^2}{2}[/tex]
 
  • #4
3,812
92
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 :grumpy:
[tex](y-y_0)=V_{0y}t+\frac{at^2}{2}[/tex]
You can take y0 as 0 since the initial displacement is zero. :smile:

Now just plug in the values and you will get the time. :)
 
  • #5
How about this then:
[tex](y-y_0)=V_{0y}t+\frac{at^2}{2}[/tex]
[tex](0-h)=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
3,812
92
That's right. :smile:

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
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 Vx and Vy right?
 
  • #8
3,812
92
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 Vx and Vy right?
Yep, that's right. :smile:

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
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(y-y_0)[/tex]

substitute in my knowns:
[tex]V_y=(0)-g(\frac{\sqrt{2h}}{g})[/tex]
[tex]V_y^2=(0)-2g(0-h)[/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
3,812
92
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(y-y_0)[/tex]

substitute in my knowns:
[tex]V_y=(0)-g(\frac{\sqrt{2h}}{g})[/tex]
[tex]V_y^2=(0)-2g(0-h)[/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?
You got the final velocity right but with a long method. :smile:
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 co-ordinates. I have never used that way. :smile:
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.
Vx=v1
Vy=[itex]\sqrt{2gh}[/itex]

Just plug in the values and your are done. :smile:

If you want to find the angle, you need to first specify the reference, Horizontal or vertical. :)
 
  • #11
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 :smile:
 
  • #12
3,812
92
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 :smile:
Your welcome!!:smile:
And remember the table approach. Whenever you deal with projectiles, the best way is to make the table :smile:

I forgot to welcome you. Welcome to the board. Hope you enjoy your stay here. :smile:
Before wandering on the board, have a look at the rules. They may help you.
 
  • #13
Thanks for the advice

Thanks again
 

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