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

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The stone is thrown horizontally from a height of 200m with an initial velocity of 10m/s. To determine its position after 2 seconds, calculate the vertical drop due to gravity and the horizontal distance traveled. The time to hit the ground can be found by solving the vertical motion equation, which gives approximately 6.3 seconds. The final speed upon impact combines both the horizontal velocity and the vertical velocity at that moment. The independent nature of x and y motions allows for straightforward calculations to find the required results.
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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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Thank you
 
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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