How Long Does a Horizontally Shot Ball Stay Airborne?

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SUMMARY

A ball shot horizontally at a speed of 10 m/s from a height of 2 meters will remain airborne for approximately 0.64 seconds before hitting the ground. The vertical motion is governed by the equation s = s0 + ut - (1/2)gt², where s0 is the initial height of 2 meters, u is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (9.81 m/s²), and t is the time in seconds. The horizontal distance traveled can be calculated by multiplying the horizontal speed by the time in the air, resulting in a horizontal distance of about 6.4 meters.

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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

[tex]s=s_o +ut-\frac{1}{2}gt^2[/tex]


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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