Saturday, May 12, 2018

HP Prime: Solving Projectile Motion with the Solve App



HP Prime:  Solving Projectile Motion with the Solve App

This blog entry is a demonstration using the Solve App to solve a variety of problems regarding projectile motion. 

Getting Started

Go to the Solver App by pressing [Apps] and selecting the Solve application.  If need be, clear all the equations by pressing [Shift] [ Esc ] and selecting (OK) at the Clear All Expressions prompt.

Enter the following equations:

E1:  X = V*COS(θ)*T
E2:  Y = V*SIN(θ)*T – 0.5*G*T^2


Variables:

X = distance in the x direction
Y = distance in the y direction
Θ = angle the projectile is shot
V = initial velocity
T = number of seconds
G = Earth’s gravitation constant:  9.80665 m/s^2, 32.174 ft/s^2

Right now, we are assuming that there is no acceleration in the X direction, the projectile starts at point (0,0), and there is no air resistance.

Finally, put the calculator in Degrees mode, which the angle indicator will display °.

Example 1:  Determining Distance and Time it takes for the Projectile to complete

Given the following data:

θ = 40°
G = 9.80665 m/s^2
V = 35 m/s
Y = 0

Select X and T to solve.  Pressing (Solve) will give the following results:

X = 0 and T = 0.  Well, that went well.  We have an obvious answer.  What we are looking for is the distance traveled after the projectile has been shot.  Fortunately, the HP Solve app allows us to provide initial guesses. 

Let’s put an initial guess of T = 100 and press (Solve) again.  We now get the following results:

X = 123.017492971
T = 4.58822662969

The projectile will travel approximately 123 m in the x direction and it will take slightly after four and half seconds before impact.



Example 2:  Determining initial velocity (V) and angle (θ)

Given the following data:

X = 125.15 m
Y = 3.10896 m
T = 5 s
G = 9.80665 m/s^2 (SI units are used)

The projectile landed on a hill 125.15 m away with an elevation 3.10896 m (10.2 ft) higher than the starting point. 

You can clear the data variables with [Shift] [Esc].  Since you are solving for V and θ this time, make sure only those variables are checked.

Select θ, give it an initial value of 0, and press (Solve). 

The results are:

θ = 45.1238194711
V = 35.474510698




Example 3:  Hitting a Moving Target

Find the required initial velocity (V) to land a projectile on a wagon moving at rate of 7.33 ft/s.  The wagon is 5.5 ft high.  The wagon is 20 ft away when the projectile is shot.  Assume the projectile is shot at 30°. 
  
The first step is to modify equation E1.  Since the X distance contains a moving object, adjust the left side to 20 + 10*T.   20 for the initial distance plus 10*T for the moving wagon. 

The equations will be set as such:

E1:   20+10*T = V*COS(θ)*T
E2:   Y = V*SIN(θ)*T – 0.5*G*T^2

Next enter the following variables:

θ = 30
Y = 5.5 (height of the wagon)
G = 32.174 (we are using US units)

Note that there only five variables this time.  Select V and T to solve for.  We want T selected because we don’t know what T is. 

Select T and enter 1 as an initial guess (actually, we really want a small positive number should do it, as long as T is not zero).  . 

We get a solution of:

T = 0.818269246898
V = 39.7700027103

This means we should fire the projectile at about 39.77 ft/s and it will land on the wagon. 




Example 4:  The Maximum Height of a Projectile

What is the maximum height of a projectile found for a projectile shot at 36° at 54 ft/s?  How far (horizontal distance) does the projectile would have traveled to get that point?

For this problem, we are going to use three equations but they will be solved separately.  Set the equations as such:

E1:  X = V*COS(θ)*T
E2:  Y = V*SIN(θ)*T – 0.5*G*T^2
E3:  0 = V*SIN(θ) – G*T

The projectile reaches its height when the rate of the change in the y direction is 0.  In calculus terms, when Y’(T) = 0 where Y’(T) = dY/dT = V*SIN(θ) – G*T.

On the Symb screen, select only E3.  Press [Num] and enter the variables:

V = 54
θ = 36
G = 32.174 (ft/s^2)

Solve for T.  The result is: T = 0.76305137487.

It takes the projectile about 0.7630 seconds for the projectile to reach its height.

The next step is press [Symb] and select the equations E1 and E2.  Now press [Num], select X and Y to solve for and press (Solve). 

The solution is:

X = 43.0981662447
Y = 15.6563253279

The distance is about 43.0982 ft and the height is 15.6563 ft.




Eddie


All original content copyright, © 2011-2018.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.  Please contact the author if you have questions.

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