Question

It is possible to shoot an arrow at a speed as high as 100m/s. If friction is neglected, how high would the arrow go? How long would it be in the air? If the bow accelerated the arrow over 0.75m, what acceleration did the arrow experience? How many g's is this?

Answer 1

The highest speed the arrow will be travelling is the moment it leaves the bow. It will then be pulled back downwards with an acceleration of 9.81 ms^2.

Since acceleration is change in velocity divided by time, and we know that the arrow at the top will be travelling at zero m/s, we know that the change in velocity is 100 - 0 (which is just 100).

To get time on it's own, we can say that time is change in velocity divided by acceleration, (100/9.81) which works out to be 10.19 seconds. Of course, now our arrow is at the top of it's journey, so it must spend the same amount of time coming back down (We ignore wind resistance and terminal velocity because you said that was ok). That gives us (10.19 x 2) = 20.38seconds.

Using the formula: s = (u+v)t/2, we can see that the height it gets to (above the bow, anyway) is ((100 + 0) x 10.19) / 2, which is 509.5 metres.

Rearranging the formula: s = (u+v)t/2, we get t = 2s/(u+v), so we know that it took (2x0.75)/(0+100) = 0.015 seconds, and we know that acceleration is change in velocity divided by time, we can see that the acceleration is (100-0)/0.015 = 6,666 ms^2, or about 679g.