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5

Air drag is given by $$F_d = \frac12 \rho v^2 A C_D$$ where $\rho$ is the density of air (variable, depending on temperature etc - 1.2 kg/m$^3$ is a reasonable approximation), $v$ is the velocity, $A$ is the projected area, and $C_D$ is the drag factor - which is a function of shape and Reynolds number. 0.5 is an OK approximation but it depends on the ...

3

I'm a bit unsure where you are unsure, but let me try. The "real" behaviour is best seen in the frame of the liquid. In this frame the bullet will shrink compared to it's stationary size, and so it will have a higher density than the water (supposing they have the same density when stationary). The acceleration downwards will be given by the difference in ...

2

you have to notice that this motion is accelerated, so if you define velocity as $d/t$ you will get the wrong result. For uniformly accelerated motion (which is your case, the acceleration is contant: g), you have to use the following relationship: $y(t)=y_0+v_0t+\frac{1}{2}at^2$ When you release the ball, $y_0=H$, $v_0=0$ and $a=-g$ so you get $... 1 You can decouple the horizontal and vertical motion of your rocket. In the vertical direction you have vertical thrust and gravity and horizontally you only have thrust (I ignore air resistance here). As you are interested in the altitude only, we only look at the vertical problem. All kinetic energy in the vertical direction is converted to potential energy ... 1 The initial kinetic energy$E_k$gets partly dissipated as friction,$E_f$, and partly converted to gravitational potential energy,$E_g$. The sum of these two must equal the original energy input, so $$E_k = E_f + E_g$$ 1 In the first part you wrote$E_{k}=E_{g}$because kinetic energy is fully converted into potential energy. But in the second part, some of the initial kinetic energy$(E_{f})$lost due to friction and part of energy left is$E_{k}-E_{f}$. Only this part is converted to potential energy$E_{g}$. Thus,$E_{g}=E_{k}-E_{f}$and this simplified as ... 1 According to WolframAlpha,$100\text{lbs}$of ice has a volume$V=0.04536 \text{m}^3$and mass$m=45.36\text{kg}$. We can find the radius of the ice ball: $$V=\frac{4}{3}\pi r^3 \rightarrow r = \sqrt[3]{\frac{3V}{4\pi}} = 0.221\text{m} = 22.1\text{cm}$$ The force of air resistance is given by the equation: $$f_\text{drag}=-\frac{1}{2}C\rho Av^2$$ Where ... 1 This type of problem can be simplified if you use the frame of reference of the elevator. Now the bolt falls from rest, chasing the "starting point" which starts a distance$h$below and moves down at a constant$6 $m/s. The bolt catches up in$3\$ seconds. Same problem, but the equations are simpler...

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