What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem
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Answer:
x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This is a projectile launch exercise, in this case we will write the equations for the x and y axes
Let's use trigonometry to find the components of the initial velocity
sin θ = / v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = vo cos θ
now let's write the equations of motion
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ + t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components