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Assuming quadratic drag, we can use the equations given at this link to answer your question. The key equations are expressed in terms of the terminal velocity $v_t$: $$v_t = \sqrt{\frac{2mg}{C_D \rho A}}$$ and the characteristic time $\tau$: $$\tau = \frac{v_t}{g}$$ Then the time to reach the peak height is $$t_{peak} = \tau \tan^{-1}\frac{v_0}{v_t}$$ ...

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Yes the bullets can fall down and injure or kill you. In fact in countries were celebratory gun firing is possible people are often injured by falling projectiles. Shooting straight up is less dangerous than at an angle because the terminal velocity is much lower than the muzzle velocity of the projectile. When shooting at an angle some of the horizontal ...

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There is always air resistance, unless you are in space. If there was no air resistance, a bullet would land at the same speed it was shot at. On Earth, a spent bullet is rarely lethal. In big wars (like World War II) it was common for soldiers to get hit by spent bullets. Normally they will hit you and fall away, causing just a bruise, An unlucky hit ...

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It is because the ball was traveling with you when you threw it. Imagine the following question: I am in a car traveling at 5m/s holding a ball. Where will the ball be relative to me in 10 seconds? Answer in my hand. To be obtuse: ball has velocity 5m/s. After 10s it will have moved forward 10 x 5 meters = 50. I have velocity 5m/s. After 10s it will have ...

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You need to use energy because the value of $g$ varies with height. So something like $\frac 12 m v^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$

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The terminal velocity is by definition the velocity for which the gravitational force is equal to the friction force of a falling body. As the friction force grow with speed (linearly for laminar flow, quadratically for turbulent flow) an object going faster that the terminal velocity will have his friction force bigger than the gravitational force and thus ...

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The mentos thrown faster than terminal velocity will experience greater drag until it once again has terminal velocity (note - terminal velocity is only ever approached asymptotically, but we can decide to call "close enough"="equal") If one thing is going faster than another for a while, after which they are traveling at the same speed, then the faster ...

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Always start with a diagram :) Sorry, I have still not solved this :( Here's my latest attempt, that demonstrates one way not to do it. Relative to the slope of the plane Projection $${v_x \choose v_y} = {u_x + a_x t_1\choose u_y + a_y t_1} \\ {s_x \choose 0} = {u_x t_1 + \frac{1}{2}a_x t_1^2\choose u_y t_1 + \frac{1}{2}a_y t_1^2}$$ where \$u_x = ...

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