What is the slope of the ground in the landing zone?

AI Thread Summary
The discussion focuses on calculating the slope of the ground in the landing zone for skiers leaving a jump at a speed of 28 m/s and an angle of 9.5 degrees below the horizontal. The horizontal distance to the landing zone is 55 m, and the skiers' landing angle with the ground is 3 degrees. The participant attempts to determine the time of flight using the horizontal motion equation but struggles with finding the x-y coordinates for the landing. There is a mention of the velocity vector aligning with the skis' direction, which is crucial for understanding the slope. The problem emphasizes the relationship between the angles and the slope of the ground at the landing zone.
Cantworkit
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[SOLVED] linear motion

Homework Statement


Skiers leave a ski jump at 28 m/s, at an angle of 9.5 degrees below the horizontal. Their landing zone is a horizontal distance of 55 m from the end of the jumo The ground is contoured at that point so that the skiers trajectories make an angle of only 3 degrees with the ground on landing. What is the slope of the ground in the landing zone?


Homework Equations


tan theta = y/x
x = x0 + v0t + 1/2gt^2
y = y0 + y0t + 1/2 gt^2

The Attempt at a Solution



t = (x - x0)/v0 = (x - x0)/v0cos9.5 = 2s
From here I can solve for the velocity vector at the landing, but I do not see how to solve for the x-y coordinates
 
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Cantworkit said:

Homework Statement



From here I can solve for the velocity vector at the landing, but I do not see how to solve for the x-y coordinates

Won't v point in the same direction as the skis? As I read the problem, the angle of the slope is three degrees less than this.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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