- #1

Radarithm

Gold Member

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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 = h

_{0}+v

_{0y}sin([itex]\theta[/itex])×t+½at

^{2}

x = x

_{0}+v

_{0x}cos([itex]\theta[/itex])×t+½at

^{2}

v

^{2}

_{x}= v

^{2}

_{0x}+ 2a(x - x

_{0})

v

^{2}

_{h}= v

^{2}

_{0h}+ 2a(x - x

_{0})

## 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}= [itex]\sqrt{2gx}[/itex] and v

_{0h}= [itex]\sqrt{2gx}[/itex]

I am obviously doing something wrong; I can't find a way to relate h, x , and g to find v

_{0}. Any help would be appreciated.

edit: x

_{0}is equal to 0.