Simple 2D Motion, rock problem

In summary, the expression for the initial speed v of the stone thrown horizontally from a height h above the ground is: v = \sqrt{2gh} where g is the constant acceleration due to gravity and h is the height from which the stone is thrown. The horizontal component of velocity remains constant while the vertical component changes due to acceleration.
  • #1

Radarithm

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Homework Statement



A stone is thrown horizontally from a height h above the ground (h > 0). The stone hits the ground after having traveled a horizontal distance x (x > 0). What is the expression for the initial speed v of the stone? Neglect air resistance. (please use the variables x and h and the constant g in your answer)

Homework Equations



h = h0+v0y sin([itex]\theta[/itex])×t+½at2
x = x0+v0x cos([itex]\theta[/itex])×t+½at2
v2x = v20x + 2a(x - x0)
v2h = v20h + 2a(x - x0)

The Attempt at a Solution



The velocity when x = 0 (after the rock hits the ground) is 0, so I used that as v in the 3rd equation. a is simply -g (gravity). After simplifying these two I got: v0x = [itex]\sqrt{2gx}[/itex] and v0h = [itex]\sqrt{2gx}[/itex]

I am obviously doing something wrong; I can't find a way to relate h, x , and g to find v0. Any help would be appreciated.

edit: x0 is equal to 0.
 
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  • #2
Show your work.
 
  • #3
v2x = v20x + 2a(x - x0x)
since v = 0 when the rock hits the ground, 0 = v20x + 2a(x - x0x)
x0 is also 0, and a is gravity, 9.8 m/s, therefore:
0 = v20x + 2gx
v20x = 2gx
v0x = [itex]\sqrt{2gx}[/itex]

and the same applies to the equation that deals with h (height)
I only just woke up so I can't really think straight :frown:
 
  • #4
Radarithm said:
v2x = v20x + 2a(x - x0x)
since v = 0 when the rock hits the ground, 0 = v20x + 2a(x - x0x)
x0 is also 0, and a is gravity, 9.8 m/s, therefore:
0 = v20x + 2gx
v20x = 2gx
v0x = [itex]\sqrt{2gx}[/itex]

and the same applies to the equation that deals with h (height)
I only just woke up so I can't really think straight :frown:

Why do you think vx is 0 ? Is there a force in x-direction which produces a deceleration such that the x-component of velocity on reaching the ground becomes zero ?
 
Last edited:
  • #5
Tanya Sharma said:
Why do you think vx is 0 ? Is there a force in x-direction which produces a deceleration such that the x-component of velocity on reaching the ground becomes zero ?

I obviously need to redo this problem.
 
  • #6
Radarithm said:
I obviously need to redo this problem.

but you need to think in correct direction...

What are the horizontal and vertical components of initial velocity (Assuming it to be 'v') ?
 
  • #7
Tanya Sharma said:
but you need to think in correct direction...

What are the horizontal and vertical components of initial velocity (Assuming it to be 'v') ?

Well the acceleration decreases in the y direction, causing the rock to hit the ground; but that's still not answering the question; I'll skip this one and come back to it later.
 
  • #8
Radarithm said:
Well the acceleration decreases in the y direction, causing the rock to hit the ground; but that's still not answering the question; I'll skip this one and come back to it later.

The acceleration remains constant in the vertical direction.It is 'g' i.e 9.8m/s2.

The velocity of the rock when it hits the ground is not zero .The horizontal component of velocity does not change during the flight.It is the vertical component of velocity which changes due to acceleration 'g'.
 

1. What is "Simple 2D Motion"?

"Simple 2D Motion" refers to the motion of an object in two dimensions, typically represented on a flat surface such as a graph or coordinate plane. It involves the use of equations and principles of physics to describe and predict the movement of an object.

2. What is the "rock problem" in relation to Simple 2D Motion?

The "rock problem" is a common scenario used to illustrate the principles of Simple 2D Motion. It involves a hypothetical situation where a rock is thrown off a cliff with a certain initial velocity and angle, and the goal is to predict its trajectory and final landing position.

3. What are the key equations used in Simple 2D Motion?

The key equations used in Simple 2D Motion are the equations of motion, which describe the relationship between an object's velocity, acceleration, and time. These equations include the equations for constant acceleration, projectile motion, and circular motion.

4. How is Simple 2D Motion different from 1D motion?

Simple 2D Motion involves the movement of an object in two dimensions, while 1D motion only considers movement in a single dimension. This means that Simple 2D Motion takes into account both vertical and horizontal components of an object's motion, while 1D motion only considers one direction.

5. What are some real-life applications of Simple 2D Motion?

Simple 2D Motion has many real-life applications, including predicting the path of a ball in sports, calculating the trajectory of a bullet, and determining the motion of planets and satellites in space. It is also used in engineering and design to analyze the movement of objects and structures.

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