Paraglider acceleration problem

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The paraglider exerts a force of 1900 N upwards on the Earth due to the lift force acting on its wings. If the lift force decreases to 1000 N, the resultant force on the paraglider becomes 900 N downwards, as the weight remains constant. To find the vertical acceleration, the relationship between resultant force and acceleration can be applied using Newton's second law, F = ma. Without knowing the mass of the paraglider, the acceleration cannot be calculated directly. Understanding these forces is crucial for analyzing the paraglider's behavior in changing lift conditions.
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A paraglider is flying horizontally at a constant speed. Assume that only two forces act on it in the vertical direction, its weight and a vertical lift force exerted on its wings by the air. The lift force has a magnitude of 1900 N. For both questions, take the upward direction to be the +y direction.
(a) What is the magnitude and direction of the force that the paraglider exerts on the earth?

This correct answer is 1900 N upwards. I need help with the next part.

(b) If the lift force should suddenly decrease to 1000 N, what would be the vertical acceleration of the glider?
m/s2
 
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If lift force decreases to 1000N, what will be the resultant force on the body now? What relationship does resultant force and acceleration have?
 
so would that mean the skydiver falls with a force of 900 N downward? How do I get his acceleration from that because i don't know his mass. Help?
 
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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