Question

A hungry bow and arrow hunter aims at a coconut on a branch. The coconut falls from the branch simultaneously as the hunter releases the arrow. Does the arrow hit the coconut. Refer to motion of a projectile.

Answer 1

Without sufficient data about the distance between the hunter and tree, speed of arrow etc. it is difficult to predict....

Answer 2

you need to know was the hunter pointing directly at the coconut, and was its speed in the x direction sufficient to get it to the coconut before it hits the ground. If both were true, it will hit the coconut because both the arrow and the coconut fall at the same rate. "Aims" is not unambiguous... a hunter doesn't assume the coconut will fall the instant he shoots...

Answer 3

Yes, it will hit the coconut - assuming that the arrow was pointed at it. I will get back with illustration and derivation.

That aside, if I were the hunter, I would aim higher than the coconut, to account for loss of altitude of arrow in flight.

Answer 4

I have attached figure. Please refer to it.

Assume that the horizontal distance of coconut is 'd' from the hunter, and at a height of 'h' from the ground. The hunter aims at the nut at angle theta from ground and releases arrow at velocity v. We are exclusively going to use equation: \[s = h _{0} + v _{0}t + 1/2 at ^{2}\]
While considering only the vertical motion of both objects (which are unaffected by any horizontal movement), h0 is 'h' for coconut and 0 for arrow. a = -g and v0 is 0 for coconut and v*sin(theta) for arrow.

The time of flight has to be calculated from the horizontal range (as its velocity is unaffected by gravity). It turns out: \[t = d/(v*\cos(\theta))\] At this time, the coconut and the arrow are above the same point on ground. If at this time, both objects are at same height above the ground, the arrow will hit the coconut.

Using the earlier equation, the height of coconut above ground: \[h _{c} = h - 1/2*g*(d/(v*\cos(\theta))^{2}\]

The height of arrow above ground is : \[h_{a}=(v*\sin(\theta))*(d/(v*\cos(\theta))) - 1/2*g*(d/(v*\cos(\theta))^{2}\]
But, \[\sin(\theta)/\cos(\theta) = h/d\] Applying this, \[h _{a} = h - 1/2*g*(d/(v*\cos(\theta))^{2}\]

So, hc and ha are the same. The arrow will meet the coconut.

Answer 5

it is possible but, with out data it is impossible to answer

Answer 6

Bottom line is that, ignoring air resistance, all objects will fall at the same rate. So the amount that the arrow falls (relative to where it was aimed) will be the same as the amount the coconut falls (relative to where it was originally). So if the arrow was aimed directly at the coconut, it will hit the coconut in its new location.

Answer 7

Since neither distance nor the speed with which the arrow is released is specifically given, the information is too sketchy to say whether the arrow will hit the coconut.

If we ASSUME that the velocity is sufficient to reach the coconut, then we can use the above explanation.

If the arrow is released with low velocity, it will not even reach the coconut, let alone hit it...

Answer 8

All the restrictions mentioned do apply. Still, I think this is a high school physics question. Its objective is to prove a point about projectile motion. It is very theoretical and doesn't need the rigorous analysis needed in engineering. It just needs a proof.

Answer 9

Both the coconut and arrow are affected by gravity (ignoring air resistance) in the same way, The arrow will fall the same amount as the coconut in the same time interval. The arrow strikes the coconut every time.