What Is the Initial Speed of a Horizontally Thrown Stone from Height h?

[Radarithm]

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+v_0y sin(\theta)×t+½at²
x = x₀+v_0x cos(\theta)×t+½at²
v²_x = v²_0x + 2a(x - x₀)
v²_h = v²_0h + 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: v_0x = \sqrt{2gx} and v_0h = \sqrt{2gx}

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.

 

[SteamKing]

Show your work.

 

[Radarithm]

v²_x = v²_0x + 2a(x - x_0x)
since v = 0 when the rock hits the ground, 0 = v²_0x + 2a(x - x_0x)
x₀ is also 0, and a is gravity, 9.8 m/s, therefore:
0 = v²_0x + 2gx
v²_0x = 2gx
v_0x = \sqrt{2gx}

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

 

[Tanya Sharma]

v²_x = v²_0x + 2a(x - x_0x)
since v = 0 when the rock hits the ground, 0 = v²_0x + 2a(x - x_0x)
x₀ is also 0, and a is gravity, 9.8 m/s, therefore:
0 = v²_0x + 2gx
v²_0x = 2gx
v_0x = \sqrt{2gx}

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

Why do you think v_x 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 ?

 

[Radarithm]

Why do you think v_x 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.

 

[Tanya Sharma]

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') ?

 

[Radarithm]

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.

 

[Tanya Sharma]

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'.