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10

Put your math aside for a minute, and take a lesson from Robert H. Goddard, in one of my all-time favorite papers. Basically your rocket consists of a payload H, and the rest of the rocket consisting of fuel mass P, plus non-fuel mass (i.e. tank) K. The secret is, as you shed P through combustion, you must also shed K. Otherwise as P gets smaller and ...


6

The answer is right there in your own math. You derived that the delta V that results from using the rocket as a single stage rocket is $$\Delta V_{\text{single stage}} = v_{ex}\ln\Bigl(\frac{M}{M-(m_{fa}+m_{fb})}\Bigr)$$ while the delta V from using the rocket as a two stage rocket is $$\Delta V_{\text{two stage}} = v_{ex} \Biggl( ...


3

Your approach is correct, now simply add indices to everything, i.e. $$y_i = v_{0,i}t_i + \frac12 a_it_i^2\quad\text{where } i\in\{1,2\}$$ and note that $t_2 = t_1 - 2\,\text{s}$. Then solve $15\,\text{m}\stackrel!=y_1 - y_2$.


2

The drift velocity is the average velocity due to an applied electric field. In a conductor, electrons scatter around at the Fermi velocity but have a net zero average (i.e., equal scattering in all directions). When the electric field is applied, the electrons are given a small velocity in one direction. Thus, we can say, $$ v_{drift}=\eta E $$ where $\eta$ ...


1

If you take the time of throwing of the first object as $0$ then the second object will start falling at $0+2$ second.Now for the second one take the time of its start of fall as 0 and so the ending time will be 2 sec less i.e$0+2 \longrightarrow 0$ and$t \longrightarrow t-2$ So for the second one $$ \frac {dx} {dt} =u+at$$ $$ dx=udt +atdt$$ $$\int^y_0 ...


1

If your position is in 3D space (which means your position vector must be defined), then there is no distinction between displacement and change in position. $s=\boldsymbol{R_f-R_i}=\Delta\boldsymbol{R}$ , where $s$ is displacement and $R$ is position. However, in $v = ds/dt$, $ds$ does not mean change in displacement but rather an infinitesimally small ...


1

The formula you have written is correct; but they are functions of time. Hence, by inserting the particular instant , say $t$ on the function ,you get the instantaneous components of velocity. Then using phythagoras theorem you will get the total instantaneous velocity. Taking your example, at time $T$ s , the X-comp. is $30$ unit and Y-comp. is $(20 - ...


1

There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.


1

The $x$ and $y$ velocities should not add to $V_0$. To understand why, imagine something moving with $V_x = 1 \frac{m}{s}$ and $V_y = -1 \frac{m}{s}$. This is something going horizontally and down; there's no reason why its velocity should be zero. The answer is that $V_0$ is the length of the velocity vector $\vec{V}$, and so it's calculated using ...


1

I would say, $$\sum \vec{F} = m\,\vec{a}_C$$ where the left hand side are the net forces applied, and $\vec{a}_C$ is the acceleration of the center of mass.



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