Kinematics 2 Problem Solutions

Each solution provided below is not necessarily the only possible solution.  If a solution differs greatly from yours, copy and paste it from your browser to your Maple worksheet, then execute it and see if the final answers differ.

Quick access to the solutions of problems (2a), (2b), (3) , (4a), (4b), (4c), and (4d).

(1a)  .41 m

Place the origin at the tip of the arrow as it leaves the bow 40 m from the target.

restart;
y[i] := 0;
v[y,i] := 0;
a[y] := -10;
y_pos := y[f] = y[i] + v[y,i]*t + (1/2)*a[y]*t^2;
x[i] := 0;
x[i] := 0;
v[x] := 140;
a[x] := 0;
x_pos := x[f] = x[i] + v[x]*t + (1/2)*a[x]*t^2;
t := fsolve( %, t );
y_pos;
y[f] := rhs(%);
miss := abs( y[f] - y[i] ); abs is the absolute value function. ?abs

(1b)

t:= 't';

Note that the following two expressions are functions of t, and that as t varies they generate the all the x and y-positions of the arrow.

> rhs(x_pos);

> rhs(y_pos);

> plot([rhs(x_pos), rhs(y_pos), t=0..4/14], labels=["x","y], title= `Robin Hood's Arrow`);      plot will not plot equations, only functions of variables.  To learn how to plot equations see ?implicitplot.

After executing this plot, place scaling = constrained into the list of plot arguments and execute the plot again.  This illustrates how you often need to try different types of plots and different parameters to achieve the best plot for a given situation. In this case scaling = constrained is a detriment.

Top

(2a)  77.5 m/s

The solution to this problem involves solving a set of four equations in four unknowns.  Depending on your approach, the solution process might be real ugly or rather straight forward.  This is true whether solving it by hand or with Maple.  The solution process involves making a series of substitutions that eliminate the number of equations and variables until you arrive at one equation in one unknown.  In theory you may make substitutions in any order you like, but for the cleanest solution the idea is to make substitutions in an order that produces the simplest expressions later on.  Learning the pitfalls of different approaches to solving large sets of equations is a matter of trial and error.

The launch speed, v, is the magnitude of the vector sum of the and (no i because the x-velocity does not change during the flight due to a lack of any forces acting along the x-axis).  Trigonometry yields the following couple of equations.

restart;
v[y,i] := v*sin(60/180*Pi);      You need to enter all angles as radians in Maple.
v[x] := v*cos(60/180*Pi);     For purposes of the discussion below, think of these two Maple assignment statements also as the two mathematical equations that created them.

Assume that both sides of the canyon are the same height, and place the origin of the coordinate system at the point of launch.  Then

y[i] := 0;
y[f] := 0;
a[y] := -10;
y_pos := y[f] = y[i] + v[y,i]*t + (1/2)*a[y]*t^2;
x[i]:= 0;
x[f] := 520;
a[x] := 0;
x_pos := x[f] = x[i] +v[x]*t + (1/2)*a[x]*t^2;

Using Maple's full evaluation to produce substitutions for and , we have reduced a set of four equations in four unknowns to a set of two equations in two unknowns.  The square of t in y_pos creates a set of equations that Maple does not solve in a very pretty manner if you place them in a set and then ask Maple to solve for the set.  Thus, in this case, stick to an assignment and a couple of solves to reach a solution for v.

t := solve(x_pos, t);
solve(y_pos,v);
evalf(%);

Top

(2b)

v := 77.5;
t := 't';

Note that the following two expressions are functions of t, and that as t varies from 0..14 s they generate the all the x and y-positions of Danger Dana during her flight across the canyon.

> rhs(x_pos);

> rhs(y_pos);

> plot([rhs(x_pos),rhs(y_pos),t=0..14], labels=["x","y], scaling=constrained, title=`Danger Dana's Canyon Jump`);      plot will not plot equations, only functions of variables.  To learn how to plot equations see ?implicitplot.

Top

(3)  58 degrees above the horizontal at an initial speed of 23 m/s.

A set of equations similar to those that solved problem (2) develop in this solution, also.  Place the origin of the coordinate axes at the point the football leaves the punter's foot.  Then

restart;
v[y,i] := v*sin(theta);
v[x] := v*cos(theta);
y[i] := 0;
y[f] := -1;
t := 4;
a[y] := -10;
y_pos := y[f] = y[i] + v[y,i]*t + (1/2)*a[y]*t^2;
x[i] := 0;
x[f] := 50;
a[x] := 0;
x_pos := x[f] = x[i] + v[x]*t + (1/2)*a[x]*t^2;
eq_set := {x_pos, y_pos};
fsolve( eq_set, { v, theta } );
subs( %, x_pos ) ;     A check.
evalf(%);

Now convert the radian angle to degrees:

evalf( 7.289713445 * 180/Pi );
evalf( % - 360 );

The plot:

theta := 7.289713445;
v := 23.37332882;
t := 't';
> plot([v*cos(theta)*t, (v*sin(theta)*t+(1/2)*a[y]*t^2), t=0..4], labels=["x","y"], scaling=constrained, title=`Punter's Punt`);

Notice that the football reaches 50 m along the x-axis just below the x-axis in this coordinate system.  Why?

Top

(4a)  .09 s and 3.95 s .

restart;
y[i]:= 1.22;
y[f]:= 3.05;
v[y,i]:= evalf(33.55*sin(37*(Pi/180)));
a[y] := -10;
y_pos := y[f] = y[i] +v[y,i]*t + (1/2)*a[y]*t^2;
solve( y_pos );
subs( t = 3.945412851, y_pos ); A check.

(4b)  2.49 m and 105.71 m

v[x] := evalf(33.55*cos(37*(Pi/180)));
a[x] := 0;
x[i] := 0;
x_pos := x[f] = x[i] + v[x]*t + (1/2)*a[x]*t^2;
subs( t = .09276595729, x_pos );
subs( t = 3.945412851, x_pos );

(4c)  No, the ball is about 1 m short along the x-axis.  For a home run it needs to be at a height of 3.05 m at any x-position greater than 106.75 m.

Top

(4d)

> plot([rhs(x_pos),rhs(y_pos),t=(.1)..(4.5)], labels=["x","y"], scaling=constrained, title=`A Little Short`);

Please send me any polite comments, suggestions, or corrections.