Speed of objects as they hit the ground

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When comparing two objects dropped from height h, one thrown downward and the other horizontally, both will hit the ground with the same speed, despite differing initial velocities. The final speed for the downward-thrown object is calculated using the formula vfy = sqrt(v0^2 + 2ah), while the horizontally-thrown object's speed combines vertical and horizontal components, resulting in vftotal = sqrt(v0^2 + 2ah). The conservation of energy principle supports this conclusion, indicating that the speeds will match upon impact. Thus, while their trajectories differ, the final speeds are equivalent. The calculations presented confirm this understanding.
yrjosmiel
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Let's say there are two objects. One is thrown directly downward at v0 and the other is thrown horizontally (90 degrees) at v0, both from height h. Which object will have a higher speed by the time they hit the ground? Would they have the same speed?

For the object thrown downward...
h = (vf2 - v02)/2a
vfy axis = sqrt(v02 + 2ah)
final speed = sqrt(v02 + 2ah)

For the object thrown horizontally
vy axis = sqrt(2ah)
vx axis = v0
add the two vectors
vftotal = sqrt(v02 + 2ah)

Is this right? Or have I miscalculated something in the way?
 
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You can also use energy conservation here.
 
Looks good to me. The matching velocities speeds make sense when you consider conservation of energy.
 
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The velocities won't be the same but the speeds will.
 
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