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The physics behind the 'shoot the bear' demo experiment, where a bullet is fired at a bear that drops from a height. The document derives the equation to determine if the bullet hits the bear and discusses the concept of projectile motion. It also addresses common questions about the experiment.
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Physics 125, Winter 2002, Prof. David Gerdes ©c2002 Regents of the University of Michigan 1
The situation is shown in Figure 1.
Figure 1: The “shoot the bear” demo. The gun is aimed directly at the bear. At the instant the bullet is fired with speed v 0 , the bear drops from a height h.
We can phrase the problem mathematically as follows: “What is the vertical position of the bear, and the vertical position of the bullet, when the bullet has moved through the horizontal distance d?” If the two vertical positions are equal, the bullet hits the bear.
To find the vertical positions of either the bullet or the bear, we need to know how long it takes the bullet to travel the horizontal distance d. This is given by
t =
d v 0 ,x
d v 0 cos θ
Now what are the vertical positions of the bear and the bullet at this time? Well, the bear starts from height h at t = 0 and drops from rest, so
ybear = h −
gt^2. (2)
The bullet starts from y = 0 and has a vertical velocity of v 0 ,y , so
ybullet = v 0 ,y t −
gt^2 = (v 0 sin θ)t −
gt^2. (3)
We need these two vertical positions to be equal at time t = d/v 0 cos θ when the bullet has traveled the horizontal distance to the bear. Comparing Equation (2) to Equation (3), we see that they differ in the first term; the − 12 gt^2 terms are common to both. So the vertical positions will be equal if and only if (v 0 sin θ)t = h. Does it? Let’s substitute our expression for t from Equation (1):
(v 0 sin θ)t = v 0 sin θ ·
d v 0 cos θ
= d
sin θ cos θ = d tan θ.
But look again at Figure 1:
tan θ =
(opp) (adj)
h d
Physics 125, Winter 2002, Prof. David Gerdes ©c2002 Regents of the University of Michigan 2
so
(v 0 sin θ)t = d tan θ = d ·
h d = h.
Therefore the vertical positions of the bullet and the bear are equal at time t, so the poor critter meets his end. Notice that v 0 canceled out, so this result is independent of the muzzle speed of the gun—provided, of course, that this speed is large enough for the bullet to reach the bear.
Here’s a helpful way to think about this experiment. Imagine that we could turn gravity off, i.e. set g = 0. In that case, the bullet travels along the straight line shown in the figure, and the bear does not fall anywhere when it lets go. So we obviously hit the bear then, for any value of v 0. When we turn gravity back on, both the bear and the bullet will “fall”. Since all objects fall at the same rate, the bullet still intercepts the bear.
Perhaps it’s confusing to think of the bullet as “falling” when it might be traveling up the whole time. Think of its “fall” as the difference between where it would have hit without gravity (the initial position of the bear, at height h), and the lower position where it actually hits. In that sense, the bear and the bullet both “fall” through the same distance.
Answers to frequently-asked questions: