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Graphing Calculator Solution for Two Dimensional Motion Problems

Your TI-83 needs to be in MODE=Degree, Par, Simul. This allows you to graph the horizontal motion (x) as a function of time (t) and to graph the vertical motion (y) as a function of time (t). Remember the physics of the problem! The horizontal motion is independent of the vertical motion and the only variable that they share is time.

Write an equation of motion for the horizontal motion of the object. Its horizontal distance(d) at any time (t)is a function of its horizontal velocity (v).

d = v cos q t Its vertical displacement (d) at any time (t) is a function of its vertical velocity (v) and the acceleration.
d = vi sin cos q t + 1/2 a t2

In our first type of problem, the object is launched from the apex of the trajectory. At this point, q = 0° and the cos q = 1 and sin q = 0.

Let's work an example of the first type of problem on the graphing calculator.

An object is projected from the top of a cliff 40 m with a horizontal velocity of 20 m/s. How long does it take to strike the ground? How far does the object travel horizontally?

Write the equation of horizontal motion. Enter it into "X1T ="

X1T=20 T
the horizontal velocity is 20 m/s. Write the equation of vertical motion. Enter it into "Y1T ="
Y1T = 0 + 1/2(-9.8)T2 + 40
it's initial vertical velocity is zero, the acceleration is that of gravity, and +40 is its initial vertical displacement.

Graph the functions. View the graph. Set Window as needed. Appropriate window settings for this problem are:

Tmin=0
Tmax=3
Tstep=0.1
Xmin=0
Xmax=60
Xscl=1
Ymin=0
Ymax=50
Yscl=1

Use the trace key to determine the maximum time is 2.86 sec and the horizontal distance is 57.14 m. You can use the "zoom in" function to get an exact answer.

Our second type of problem differs in that the object is projected from the horizontal and has a value for the angle q. Also, the initial vertical displacement is zero.

Let's work an example.

An object is projected with a velocity of 20 m/s at an angle of 37X1T ="

X1T=20 cos 37 T
the horizontal velocity is 20 cos 37 Write the equation of vertical motion. Enter it into "Y1T ="
"Y1T = 20 sin 37 + 1/2(-9.8)T2
it's initial vertical velocity is 20 sin 37, the acceleration is that of gravity, and the initial vertical displacement is zero.

Graph the functions. View the graph. Set Window as needed. Appropriate window settings for this problem are:

Tmin=0
Tmax=3
Tstep=0.1
Xmin=0
Xmax=50
Xscl=1
Ymin=0
Ymax=10
Yscl=1

Use the trace key to determine the maximum time is 2.86 sec, the horizontal distance is 39.29 m, and the maximum vertical displacement is 7.40 m. You can use the "zoom in" function to get an exact answer. You can determine the vertical velocity at any time t by using the dy/dt function found under 2nd calc. Enter twice for the answer.