Motion in two dimensions please?

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To determine the horizontal displacement of the hammer, the time of flight must first be calculated using the vertical motion equation, Y = Voy*t + (g*t^2)/2, where Y is -10 m. Given the initial vertical speed (Voy) can be derived from the hammer's initial speed of 4 m/s at a 30-degree angle, the vertical component can be calculated. Once the time (t) is found, it can be substituted into the horizontal displacement equation, X = Vox*t, where Vox is the horizontal component of the initial speed. This process will yield the total horizontal distance the hammer travels before hitting the ground. The calculations involve solving a quadratic equation for t based on the vertical motion.
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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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