Motion on a Plane: Solving Problems with Stone Thrown from 200m Tower

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SUMMARY

The problem involves a stone thrown horizontally from a height of 200 meters with an initial velocity of 10 m/s. The stone's motion can be analyzed by separating the vertical and horizontal components. After 2 seconds, the stone's horizontal position is 20 meters, and it will hit the ground after approximately 6.4 seconds. Upon impact, the stone's speed will be approximately 34.64 m/s, calculated by combining the horizontal and vertical velocity components at the moment of impact.

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Question:
A stone is throws with an initial horizontal velocity of 10m/s form the top of a tower 200m high.
a)where is the stone 2 seconds after being thrown?
b)when will it hit the ground?
c)what is the speed upon hitting the ground?

(this is not my homework it a question that wasn't discussed in class and i am not able to answer it)
 
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The trick is that the x and y motions are independent and are superimposed on each other (an approximation good for the world being flat). If it is thrown horizontally, then the initial velocity has no y component. First, solve a problem as thought there is only y-motion, that it, a stone dropped vertically from rest. Find the time to hit the ground, the position at 2 seconds, and the final speed just before it hits the ground. Then do another problem as if there is only x-motion, as if it were sliding on a table. Neglecting air resistance, x-motion has a constant horizontal velocity, because gravity doesn't affect the x-motion. Find the position at 2 seconds. Find the x-velocity at the time when the first part of the problem, the y-solution, said it will hit the ground. Now you can combine the x-results and the y-results to get the things they ask for.
 
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