Magnitude of the overall Velocity problem

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To determine the magnitude of the overall velocity of a package dropped from a plane at a height of 3230 m with a horizontal velocity of 110 m/s, the time of fall is calculated to be 25.7 seconds using the equation d = vo(t) + 1/2at^2. The vertical component of the final velocity (vfY) is derived from the initial vertical velocity (voY = 0) and the acceleration due to gravity (aY = -9.8 m/s²). The overall velocity is a vector sum of the horizontal and vertical components, requiring the use of Pythagorean theorem to combine these velocities. The final calculation yields the magnitude of the overall velocity at the moment the package touches the ground.
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1. A plane drops a package from a height of 3230 m. The plane's horizontal velocity was 110 m/s at the instant the package was dropped. What is the magnitude of the overall velocity of the hamper at the instant it touches the ground?



2. d=vo(t) + 1/2at^2



3. dx
vx 110
t 25.7

vfY
voY 0
aY -9.8
dY -3230
t 25.7

I found the time from the equation above. How would you find the overall velocity, and then how would you find the magnitude of the overall velocity?
 
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Calculate the final velocity by using the fact that the box is experiencing a constant acceleration. To find the 'overall' velocity, note velocity's vector nature.
 
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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