How Long Does a Horizontally Shot Ball Stay Airborne?

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A ball shot horizontally at 10 m/s from a height of 2 meters will remain airborne until it hits the ground. To determine the time in the air, the vertical motion must be analyzed using the equation s = s0 + ut - (1/2)gt^2, where s0 is the initial height of 2 meters. The final vertical position (s) is 0 when the ball hits the ground, allowing for the calculation of time (t). The horizontal distance traveled can then be calculated by multiplying the horizontal speed by the time in the air. Understanding both the vertical and horizontal components is essential for solving the problem accurately.
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Homework Statement


A ball is shot horizontally at a speed of 10 m/s from a height of 2 meters. How long is the ball in the air and how far horizontally does it travel before it hits the ground.


Homework Equations


V2 Equation


The Attempt at a Solution


I am assuming I need to find both the X and Y components right? I know that the final Y velocity must be 0 because it stops at the top of its trajectory. What is throwing me off is the 2 meters it starts at as well as how I am going to get the horizontal distance.
 
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IneedHelp:-) said:

Homework Equations


V2 Equation

s=s_o +ut-\frac{1}{2}gt^2


where s=displacement and s0= initial displacement.

So considering vertically, what is s0? and you want to find the 't' for which s=0.
 
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