If I throw an 18g dart from a height of 70" to a target that is 77" away & 68" high, and it takes .24 seconds to get there. Approx how heavy a dart of similar design would be required to hit the target if it's 3" further away using the same arm tempo?

If Acceleration is a necessary component to figuring this out, let's assume the dart leaves my hand traveling 25.1 feet per second and after traveling 4 ft the speed has dropped to 23.5 feet per second.

Dec 26, 2016

The range of a simple projectile is mass independent.

Explanation:

In simple projectile motion problems, vertical acceleration will always be equal to $- g$, which on Earth is $- 9.8 \frac{m}{s} ^ 2$ (constant), where the horizontal acceleration will always be zero (constant). A projectile is an object with an initial non-zero, horizontal velocity whose acceleration is due to gravity alone. As soon as the dart leaves your hand, it becomes a projectile, and has only vertical acceleration due to gravity. This is not to say that other forces do not act on the object (e.g. air resistance), just that the effect of these forces are (typically) minimal in comparison.

With that being said, we know that all falling objects will accelerate at the same rate under the influence of gravity. Of course, we do not observe this when an object is light enough that air resistance is not negligible, such as in dropping a feather. However, we know that if we were to drop a feather and a brick at the same time from the same height, for example, in a vacuum (where there is no air resistance), both would then fall at the same rate.

This tells us that ( ignoring air resistance ), the mass of the projectile is irrelevant. Now, in the case of a dart, one can make the argument that air resistance is not negligible, but at that point, this becomes something far beyond an introductory physics level, 2-dimensional projectile motion problem, but rather one requiring more complex mathematics involving differential equations as you account for air resistance. An object ceases to be a projectile once anything is done in an attempt to change its trajectory, and thus the only relevant quantities that might vary between projectiles are initial velocity and position to produce differences in flight time, distance traveled, etc.

The kinematic equation for the horizontal range of the projectile is given by:

$\Delta x = {v}_{i x} \Delta t$

We can see that the horizontal distance traveled depends only on the initial horizontal velocity and fall time. If you attempted to use energy conservation, you would find that the mass term cancels. As long as the darts are thrown with the same initial velocities, they will travel the same distance.

Again, the above assumes a problem of simple projectile motion. In the real-world situation we cannot ignore outside forces, such as air resistance. These forces have significant effect on the flight of darts and produce changes in their motion which are not accounted for by simple kinematics.

Since we're only adding 3 inches, the additional mass needed is quite small - an increase of $\frac{1}{4}$g. That said, I have no idea if I've ignored something that makes this a ridiculous result.

Explanation:

Before I dive in with my answer, I'd first like to say that it is real-world problems that highlight the two main branches in physics - the theorists (who predict results that others then confirm through experimentation) and the experimentalists (who find results through experiments that others are then left to find the explanation of what was found). It seems to me that in this case it'd be far easier to experiment than theorize - but as I'm not a darts person, I'm left with theorizing!

We are given the weight and initial speed of a dart, also the final speed at impact with the target a certain distance away. The target is then moved and we are being asked for an adjustment to the dart's mass so as to make it travel the additional distance without adjusting the launch.

Some things I'm going to assume:

• the force of gravity, because of the negligible amount of time the dart is in the air, can be ignored.

• the path the dart will travel will be essentially a straight line from the point of launch to the point of impact

• the arm of the thrower will be moving at the same speed, regardless of the weight of the dart. This will give the darts the same speed at launch. Any difference in velocity (so the direction the dart is being thrown) will be essentially the same at the shorter and longer distance

Let's now talk first about the first dart throw (at the original distance).

We know the dart leaves the thrower's hand at a kinetic energy of:

$K E = \left(\frac{1}{2}\right) m {v}^{2} = \left(\frac{1}{2}\right) 18 {\left(25.1\right)}^{2} = 5670 \left(g {m}^{2} / {s}^{2}\right)$

and we know the KE at a distance of 4 feet to be:

$K E = \left(\frac{1}{2}\right) m {v}^{2} = \left(\frac{1}{2}\right) 18 {\left(23.5\right)}^{2} = 4970 \left(g {m}^{2} / {s}^{2}\right)$

We have the distance being traveled by the dart to be the hypotenuse of a triangle 77 inches long and 2 inches deep, so it's:

$\sqrt{{77}^{2} + {2}^{2}} = 77.02$ so we can ignore the idea of needing to figure in the drop from the arm to the target and instead focus on this as being a horizontal straight line.

The energy loss due to air friction and whatnot at the 4 ft mark is:

$5670 - 4970 = 700$ and as a percentage of the initial energy, is:

700/5670~=12.5% at the $\frac{48}{77} \cong .6$ distance. We can assume then that there is a total energy loss of:

12.5/.6~=21%, which means the dart impacts the target with an energy of:

$5670 \times \left(1 - .21\right) \cong 4480 \left(g {m}^{2} / {s}^{2}\right)$

I'm going to assume we want the heavier dart to impact the target with the same energy.

Ok - so we're adding 3 inches to the distance. What mass is needed for the dart?

The distance increases from 77 to 80, which is a change of

3/77~=4%

which means the energy loss during flight will be

21%(1.04)~=22%

So that means we want the dart to have an initial energy of:

$\frac{4480}{1 - .22} \cong 5744 \left(g {m}^{2} / {s}^{2}\right)$

Which can be achieved this way:

$K E = 5744 = \left(\frac{1}{2}\right) m {v}^{2} = \left(\frac{1}{2}\right) m {\left(25.1\right)}^{2} \implies m \cong 18.25 g$

So if I've got everything assumed and calculated correctly, by adding a tiny bit of weight, namely $\frac{1}{4} g$, you'll achieve the target without a need to change the arm swing.

Now it's time for someone else to go off and validate (or refute) this theory with some actual results!