Motion in two dimensions please?

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SUMMARY

The discussion focuses on calculating the horizontal displacement of a hammer dropped from a roof at a speed of 4 m/s, with the roof inclined at a 30-degree angle and 10 meters above the ground. The equations of motion used include Y = Voy*t + (g*t^2)/2 for vertical displacement and X = Vox*t for horizontal displacement. To find the time (t) until the hammer hits the ground, participants suggest setting Y to -10 m and solving for t using the known initial vertical velocity (Voy). This approach leads to determining the horizontal distance traveled by the hammer.

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A worker works in a roof and let's a hammer fall over the roof with the speed 4m/s.The roof forms an angle 30 degree related to the horizon and its lowest point its 10meters from the ground.What is the horizontal displacement of the hammer from the moment he leaves the roof to the moment he touches the ground.

So X=Vox*t
I need to find that t.
Y=Voy*t +g*t^2/2

Y is 10 m ...what do I do now?
 
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Elaia06 said:
A worker works in a roof and let's a hammer fall over the roof with the speed 4m/s.The roof forms an angle 30 degree related to the horizon and its lowest point its 10meters from the ground.What is the horizontal displacement of the hammer from the moment he leaves the roof to the moment he touches the ground.

So X=Vox*t
I need to find that t.
Y=Voy*t +g*t^2/2

Y is 10 m ...what do I do now?

When t = 0, Y = 0. Let Y = -10m and solve for t. You know what Voy is.
 

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