# If a stone is horizontally launched at a certain height how do you describe the magnitude of its vertical velocity as it approaches the ground?

**Question: If a stone is horizontally launched at a certain height how do you describe the magnitude of its vertical velocity as it approaches the ground?**

When a stone is horizontally launched from a certain height, its vertical velocity increases due to the acceleration caused by gravity. Assuming there is no air resistance, the only force acting on the stone in the vertical direction is gravity, which accelerates the stone at approximately 9.81 m/s² towards the ground. The magnitude of the stone's vertical velocity as it approaches the ground can be described by the equation \( v = gt \), where \( v \) is the vertical velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time elapsed since the stone was launched. As the stone falls, its vertical velocity will increase linearly over time, reaching its maximum just before impact. This final velocity can also be calculated using the equation \( v = \sqrt{2gh} \), where \( h \) is the height from which the stone was launched.

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