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Let's imagine commonly seen in movies and video games "rain of arrows". Archers shoot up to hit enemies behind the obstacles, for example.

As Brittanica states:

“The best longbows were made of yew, might have required a force of as much as $150$ to $180$ pounds ($70$ to $80$ kg) to draw, and shot arrows a cloth yard (about $37$ inches, or $94$ cm) long, with an effective range of some $450$ to $1,000$ feet ($140$ to $300$ meters) depending on the weight of the arrow.”

Is it enough data to determine effectiveness of rain of arrows?

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  • $\begingroup$ They were effective enough to terrify every other army in Europe. There is one reputable eye witness account of a knight being shot in the leg. The arrow went through his metal chain mail, through his leg, through the horse's leather saddle, and killed the horse. Even a complete beginner can shoot an arrow right through a human-sized piece of meat, at 30 or 40 yards range. $\endgroup$ – alephzero Aug 16 at 21:30
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I'll take the lowest values as this is a question about effectiveness, so let's assume no air friction and that the bow doesn't dissipate energy in any other form than pure kinetic energy. The force felt was 70kg so converting that to newtons gives us 686.70 newtons dragging the 0.94m long arrow back averaging 343.35 (correction by alephzero) thus giving us a total energy of about 323 Joules stored in the bow. let's assume the archer knows what he's doing and firing at a 45° angle (as it goes the farthest at that angle) the equation for vertical motion is then: $y = v_{y}*t - \frac{1}{2}*g*t^{2}$ with the second zero of y (when it hits the ground) at time: $\frac{-v_{y}+sqrt(v_{y}^{2} + 2*g)}{g}$ (which you convert to v using $v_{y} = \frac{sqrt(2)*v}{2}$ this is also the time it takes for the arrow to reach (at minimum) 140 metres and thus $v_{x} = sin(45°)*v$ gives that $140 = v_{x}*t = \frac{sqrt(2)*v}{2}$ from which we get a velocity of about 30 m/s (plotted the function and found the intersection). with this then we can calculate the mass of the arrow: $T (= kinetic energy) = \frac{1}{2}*m*v^{2}$ (=earlier potential energy stored) which gives a mass of 0.7kg. So in the end we have an arrow of 0.7kg traveling at 30m/s to someone's face.

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    $\begingroup$ Note that you don't really need to calculate the trajectory or the time of flight. If you're firing on level ground and neglecting air resistance, then the kinetic energy when the arrow leaves the bow is the same as when it lands. $\endgroup$ – Michael Seifert Aug 16 at 21:03
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    $\begingroup$ Your energy is wrong by a factor of 2 because the average force to pull back the bow is only half the maximum. (Actually it's not exactly half for a real bow, but 1/2 is a better approximation than 1). $\endgroup$ – alephzero Aug 16 at 21:26

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