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38

Other answers don't mention the fact that no single impulse (e.g, like being fired from a gun) can launch a projectile into orbit. A purely ballistic projectile fired from a gun must either crash back into the planet, or it must escape from the planet altogether. In order to achieve orbit, at least two impulses must be applied to the projectile. The first ...


23

Anything launched into orbit by such a gun needs to travel at orbital velocity (in fact above orbital velocity) in the lower atmosphere. That's generally undesirable, to put it mildly: there will be really serious heating.


12

Aside from the interior ballistic aspects of these various projects, it was quickly realized that any satellites launched by gun would have to withstand high g-loadings during firing of the gun and the size and mass of the satellite would be greatly constrained by the dimensions of the bore of the gun and the maximum impulse which could be provided by the ...


8

I think the heart of the question is whether one could arrange a continuous combustion of propellant along the length of the barrel. In that way the acceleration occurs along the length of the barrel in a more gentle way. Since the expanding gases from the propellant in a shell casing expand and the pressure of the expanding gases declines along the way it ...


3

If I understand your notation right, then by $h_L$ you mean the height of the left weight? In that case your formula doesn't make sense: To account for energy conservation, you would have to take the heights before and afterwards of the right weight. Then you get \begin{align} m_R*g*h_{before} = m_R*g*h_{after} + \frac{1}{2} m_L * v^2 \end{align} You ...


3

These are a couple of classical equations of motion derived from Newton's laws, dealing with the motion of a body with initial velocity of 50 m/s at an angle of $\theta$ degrees (or radians) with respect to the horizontal/ground/$x$-axis, assuming that the coordinate system is flat. The object is subject to a downwards acceleration, which we infer is due to ...


2

Let the velocity at instant be v(vector) = (Vx) i + (Vy) j. i and j denote unit vectors, along x and y axis respectively. dv/dt = a(acceleration) = - gj. dv/dt = [(Vx(final) -Vx(initial))i + (Vy(final) - Vy(initial))j]/dt = - gj. By initial and final I mean Vx and Vy at time t and t + dt. As the resulatant is only along the y axis the X component must be 0 ...


1

The equations look like the displacement of an object subject to zero acceleration (first equation) or the acceleration of gravity (second equation). If we assume that the first equation represents horizontal motion and the second, vertical motion (suggested by the 9.80 m/s$^2$), we can try to come up with a scenario. By inspection of equation (1) we see ...


1

The equation of motion for the ball from the time it bounces till the time it hits the ground again is $$ y = v_0t - \frac{1}{2}at^2 $$ where ground level is $y=0$, and $v_0$ is the velocity going up after adjusting for the coefficient of restitution, and $t$ is the time since the bounce. This equation will take the ball through its peak and back to the ...


1

Your formula's wrong. You've got $v=\frac12 at^2$, whereas that's the formula for $y$=height. Velocity's actually $v=at$ (with $a=9.8\mbox{m/sec}^2$).


1

What you have read is only valid in a vacuum. With air resistance the drag is a function of the total velocity, so in reality the deceleration on each axis also depends on the other.


1

If we assume that $g$ (acceleration due to gravity) is constant and parallel to our $y$ axis (but in the opposite direction), then we have: (Initial velocity=$v_0$, Launch angle=$\theta$, Initial position=$(x_0,y_0)$) $$a_y=-g\qquad (1)$$ $$\Longrightarrow \; v_y=-gt+v_0\sin \theta\qquad (2)$$ $$\Longrightarrow \; y=-\frac 12 gt^2+v_0 \sin \theta\; ...



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