Two-Dimensional Motion Concepts Applied

It's not too hard to believe that objects under the influence of gravity fall at the same rate independent of whether they are moving horizontally or not. Galileo's thought experiment about the cannonball on the boat mast seems reasonable after all. The parabolic shape is also not unfamiliar to anyone who's tried to visualize the shape of a falling object with an initially horizontal velocity. But is the answer so easily visualized if the object starts off with both a horizontal and a vertical velocity? According to Galileo's prescription, we should assume no difference in how we approach the problem. Horizontal and vertical motion are completely independent. One way to see that this still works is to work out quantitatively what happens for the hunter-monkey demonstration.

To remind you: a hunter fires a dartgun with a harmless sedative at a monkey hanging from a vine a distance h vertically above the dartgun and a distance d horizontally away from the dartgun. The hunter aims directly at the monkey and fires, but just as the hunter fires, the monkey, using its incredible spider-monkey sense, realizes what's up and drops from the vine. Does the monkey avoid the dart?

Solution: Our approach is standard. We draw a picture, then analyze the words in the problem to come up with the values for the variables we define. First, we need a coordinate system. Let's define up and to the right as positive directions with the origin being the end of the dartgun.

It's almost always useful to find the components of the acceleration and initial velocity. The acceleration is easy since only gravity is acting to change the motion, so a_x = 0 and a_y = - g. For the initial velocity, we have

v_0x = v₀ cos(theta)
v_0y = v₀ sin(theta)

The equations for motion for the monkey and the dart are:

	dart	monkey
x:	xdart = v0xt	xmonkey = d
y:	ydart = v0yt - ½ gt2	ymonkey = h - ½ gt2

Assuming that the dart does hit the monkey, the condition for a collision of dart and monkey is for x_dart = x_monkey and y_dart = y_monkey for some time t, i.e. we want to see if it's possible to find

v0cos(theta)*t	= d
v0sin(theta)*t - ½ gt2	= h - ½ gt2 ==>
v0sin(theta)*t	= h

If we divide the second equation by the first, then

(v₀sin(theta)*t)/ (v₀cos(theta)*t) = h/d ==>
tan(theta) = h/d

Therefore, for some time t, we will always find a solution which yields a collision, provided tan(theta) = h/d. However, if we go back to the wording in the original problem we find that the definition of "hunter aims directly at the monkey" is precisely that tan(theta) = h/d so we are guaranteed a "hit" if the monkey doesn't reach the ground before the dart gets to it. Note that with the exception of hitting the ground, the "collision" of dart and monkey is inevitable no matter what value of v₀ we use. That is to say, the condition for collision does not depend on the magnitude of v₀, just its direction.

Suppose we wanted to know the time of the collision between dart and monkey; what would we need to do? The consideration of this is in the Maple file along with the solutions to the problems posed below.

We've seen the demonstration of the hunter-monkey experiment so you know that the solution given above does in fact work. You don't need to pick any particular initial speed to be sure that the monkey gets hit, you just need to aim right at the monkey.

Let's do something a little more practical as long as Maple is available to do the algebra for us. Consider the following problems:

1. A baseball player hits a ball so that it's initial velocity is 40 m/s at an angle of 63° w.r.t. the horizontal. The ball leaves the bat at a height 1 meter above the ground and heads for a wall 3 meters high, located 100 meters from the batter.
   1. Does the batter get a homerun or is the ball likely to be caught for an easy out?
   2. What's the smallest speed with which the batter can hit the ball at 63° w.r.t. the horizontal and still make it over the wall?

   Check the Maple file for the solutions.

2. An olympic skeet shooter wishes to shoot a clay target that flies through the air. The target leaves from ground level 10 meters from the shooter with a velocity of 60 m/s directed 45° to the horizontal. The shooter fires a bullet with a velocity of 400 m/s directed at angle theta w.r.t. the horizontal. The bullet leaves the gun 2 meters above ground level. If the shooter fires 1 second after the clay target is fired, at what angle theta would he/she have to aim and how soon after firing would the bullet hit the target?

   Again, check the Maple file for solutions.

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