01-07-2017, 05:32 PM

HP 12C: Projectile Motion, No Air Resistance: Maximum Distance (U.S. Units)

For an object that travels in a projectile motion, we can track its range (distance traveled from the beginning) and height by:

R = v^2 * sin (2 * θ)/g

H = (v^2 * (sin θ)^2) / (2 * g)

Where:

v = initial velocity

θ = initial angle

g = Earth’s gravity. For in US units, g = 32.1740468 ft/s^2.

This program uses the approximation g ≈ 32.174 ft/s^2

The projectile will have maximum range (distance) if we aim the object at 45°.

Aside: Why?

Let’s let range (R) be a function of angle (θ):

R = v^2/g * sin(2 * θ)

Find the critical points by finding the zero of the first derivative:

dR/dθ = 2 * v^2/g * cos (2 * θ)

0 = 2 * v^2/g * cos (2 * θ)

0 = cos (2 * θ)

arccos 0 = 2 * θ

π/2 = 2 * θ

θ = π/4

Now we can use the second derivative to test whether the function is at a maximum (less than 0) and minimum (more than 0) at the crucial point.

d^2 R/dθ^2 = -4 * v^2/g * sin(2 * θ)

Let θ = π/4

-4 * v^2/g * sin(2 * π/4) = -4 * v^2/g * sin(π/2) = -4 * v^2/g < 0

(We are assuming the initial velocity is positive, and g ≈ 32.174 >0)

Since the second derivative at θ = π/4 is negative, the range is at its maximum.

Note that calculus is done when the angles are measured in radians. π/2 in degrees is 90° and π/4 in degrees is 45°. (We are only concentrating on angles between 0° and 90°)

To find the maximum range and height, substitute at θ = 45° and range and height are:

R = v^2 /g

H = v^2 / (4 * g)

The time this certain projectile lasts is:

T = (v * √2) / (2 * g)

Program:

Keep in mind: this is done on the HP 12C (regular). For the HP 12C Platinum, the code for Last X is 43, 40 ([ g ] [ + ])

STEP; KEY; CODE NUMBER

01; STO 1; 44, 1

02; 2; 2

03; ÷; 10

04; LST x; 43, 36

05; √x; 43, 21

06; *; 20

07; 3; 3

08; 2; 2

09; .; 48

10; 1; 1

11; 7; 7

12; 4; 4

13; STO 0; 44, 0

14; ÷; 10

15; R/S; 31

16; RCL 1; 45, 1

17; ENTER; 36

18; *; 20

19; RCL 0; 45, 0

20; ÷; 10

21; R/S; 31

22; 4; 4

23; ÷; 10

24; GTO 00; 43, 33, 00

Input: velocity in ft/s (convert from mph to ft/s by multiplying it by 22/15)

Output:

time of projectile in seconds, [R/S]

range of projectile in feet, [R/S]

height of projectile in feet

Example:

V = 25 mph = 36.6666667 ft/s (110/3)

Output:

Time: 0.81 sec, Range: 41.79 ft, Height: 10.45 ft

Source: Rosenstein, Morton. Computing With the Scientific Calculator Casio: Tokyo, Japan. 1986. ISBN-10: 1124161430

For an object that travels in a projectile motion, we can track its range (distance traveled from the beginning) and height by:

R = v^2 * sin (2 * θ)/g

H = (v^2 * (sin θ)^2) / (2 * g)

Where:

v = initial velocity

θ = initial angle

g = Earth’s gravity. For in US units, g = 32.1740468 ft/s^2.

This program uses the approximation g ≈ 32.174 ft/s^2

The projectile will have maximum range (distance) if we aim the object at 45°.

Aside: Why?

Let’s let range (R) be a function of angle (θ):

R = v^2/g * sin(2 * θ)

Find the critical points by finding the zero of the first derivative:

dR/dθ = 2 * v^2/g * cos (2 * θ)

0 = 2 * v^2/g * cos (2 * θ)

0 = cos (2 * θ)

arccos 0 = 2 * θ

π/2 = 2 * θ

θ = π/4

Now we can use the second derivative to test whether the function is at a maximum (less than 0) and minimum (more than 0) at the crucial point.

d^2 R/dθ^2 = -4 * v^2/g * sin(2 * θ)

Let θ = π/4

-4 * v^2/g * sin(2 * π/4) = -4 * v^2/g * sin(π/2) = -4 * v^2/g < 0

(We are assuming the initial velocity is positive, and g ≈ 32.174 >0)

Since the second derivative at θ = π/4 is negative, the range is at its maximum.

Note that calculus is done when the angles are measured in radians. π/2 in degrees is 90° and π/4 in degrees is 45°. (We are only concentrating on angles between 0° and 90°)

To find the maximum range and height, substitute at θ = 45° and range and height are:

R = v^2 /g

H = v^2 / (4 * g)

The time this certain projectile lasts is:

T = (v * √2) / (2 * g)

Program:

Keep in mind: this is done on the HP 12C (regular). For the HP 12C Platinum, the code for Last X is 43, 40 ([ g ] [ + ])

STEP; KEY; CODE NUMBER

01; STO 1; 44, 1

02; 2; 2

03; ÷; 10

04; LST x; 43, 36

05; √x; 43, 21

06; *; 20

07; 3; 3

08; 2; 2

09; .; 48

10; 1; 1

11; 7; 7

12; 4; 4

13; STO 0; 44, 0

14; ÷; 10

15; R/S; 31

16; RCL 1; 45, 1

17; ENTER; 36

18; *; 20

19; RCL 0; 45, 0

20; ÷; 10

21; R/S; 31

22; 4; 4

23; ÷; 10

24; GTO 00; 43, 33, 00

Input: velocity in ft/s (convert from mph to ft/s by multiplying it by 22/15)

Output:

time of projectile in seconds, [R/S]

range of projectile in feet, [R/S]

height of projectile in feet

Example:

V = 25 mph = 36.6666667 ft/s (110/3)

Output:

Time: 0.81 sec, Range: 41.79 ft, Height: 10.45 ft

Source: Rosenstein, Morton. Computing With the Scientific Calculator Casio: Tokyo, Japan. 1986. ISBN-10: 1124161430