Trajectory Problems

Baldwin

Trajectory problems involve the parameterization of an object’s motion into its horizontal movement and its vertical movement separate from each other It requires us to do three things that are slightly not true. First, let’s assume the Earth is flat. Second, let’s assume there is no air. And third, let’s assume that the object is a singularity with no size. This allows the object to follow a parabolic path without influences that actually do exist, but will alter the problem to more difficult conditions.

A typical problem would be: An object is thrown from a 325 foot tall building at an initial velocity of 86 feet per second at an upward angle of 39^o.

There exists three terms of each vector’s parametric equation; the acceleration term, the initial velocity term, and the initial position term. For the vertical parametric component, the acceleration term is a t² term whose coefficient is half the Earth’s acceleration, which is 32 feet per second squared downward. So the term is -16t². If we were in metric mode, it would be 9.8 meters per second squared making the first term -4.9t². The second term is the initial velocity term, which would be V_osinqt, where V_o is the initial velocity and q is the initial angle. For this particular problem, that would be 86 feet per second times the sine of 39^o, which is about 54.1t. The third term is the initial height, or h_o, which is 325 feet. So for the vertical component, we have y(t) = (a/2)t² + V_osint + h_o, or y(t) = -16t² + 54.1t + 325. For the horizontal parametric component, our acceleration term is 0 since there is no acceleration or deceleration (absent of air). The second term is V_o times the cosine of the initial angle, or V_ocosq, and for this problem that’s 86 feet per second times cosine 39^o or about 66.8t. The third term is the initial position in the horizontal sense, and if we establish the coordinate axis so that the origin is directly below the launching point, then it is also 0. So there is only one term in the horizontal component, making x(t) = V_ocosqt, or x(t) = 66.8t.

There are generally 8 questions in a trajectory problem, and a few more once you’re in calculus. Here they are.

When did the object hit the ground? If you look at the vertical component, it is accounting for the height of the object. When the height is zero, it is at the ground. Since it is a quadratic equation, then soling the quadratic with the quadratic formula will account for an elevation of zero. There will be two answers usually, one when the object hits the ground, and another that is not part of the problem situation. For our problem, the two solutions to the quadratic formula are -3,12 seconds and 6.50 seconds. Since -3.12 second occurred before the object was launched, it is wrong, leaving us with 6.50 seconds.

What was the object’s range? The range is the distance from the impact point and the point on the ground directly below the launching point. The horizontal component tells us how fast this object travels each second, so injecting the impact time into this equation will tell us how for it was when it hit. For our problem, that would be about 66.8 times 6.50, or 434 feet. Keep in mind that I am rounding answers off and errors may accumulate.

How high did the object get? If you look at the vertical component, you will see that it is a quadratic equation comparing time with height. The vertex of this parabola is V = [-b/2a , c – b²/2a ]. Co the height at the vertex is c – b²/2a, or 325 – 54.1²/(-64) = 370.7 feet.

When did the object get to its highest point? This would be the first component of the vertex formula for the time/height parabola, or –b/2a, which is -54.1/(-32) = 1.69 seconds.

How far downrange was the object when it was at the highest point? This would be the horizontal component’s equation when time was –b/2a. Using our problem, that would be 66.8 times 1.69 equals 112.9 feet.

What was the object’s vertical impact velocity? One of the things we will learn in calculus is the derivative. The derivative of position will be velocity, and the derivative of velocity will be acceleration. For each term in a polynomial, the derivative will the coefficient multiplied by the exponent times the variable to one power less than it is now. So for the -16t² term, the derivative will have a coefficient of -16 times 2, or -32, and the variable will have an exponent of 2 minus 1, or 1. The second term will have a coefficient of 54.1 times 1, and the variable’s exponent will be 1 minus 1 is 0. The third term will have a coefficient of 325 times 0 is 0, which makes it go away. That leaves the velocity for the vertical component becoming V_y = -32t + 54.1. IF we plug in to t the impact time, then we will have the vertical impact velocity, which would be -32 times 6.5 plus 54.1 = -153.9 feet per second. The negative sign means that it is downward in direction.

What is the object’s absolute impact speed? This combines the vertical component’s velocity with the horizontal component’s velocity to get an absolute speed, which is the hypotenuse of the right triangle whose legs are the two velocities. That will give us A.I.S. = 167.8 feet per second.

What is the object’s impact angle? The impact angle will have a vertical component and a horizontal component, which divided gives us the tangent of this angle. So the angle will be the inverse tangent of this ratio. Our impact angle f=tan-1[153.9/66.8]=66.^o5.

In AP Calculus AB we will discuss the length of the curve in which the object travelled, which will be our ninth question and answer. I hope this was helpful in some way.

Baldwin