The speed of an arrow fired from a compound
bow is about 79 m/s. An archer sits astride
his horse and launches an arrow into the air,
elevating the bow at an angle of 39◦
above the
horizontal and 2.4 m above the ground.
A.) What is the arrow’s range? The acceleration of gravity is 9.81 m/s^2. Assume the ground is level and ignore air resistance. Answer in units of m.
B.) Now assume his horse is at full gallop, moving in the same direction as he will fire the arrow, and that he elevates the bow the same way as before. What is the range of the arrow at this time if the horse’s speed is 13 m/s? Answer in units of m

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

Equations

X axis:

[tex]X=v_{ox}*t[/tex]

[tex]v_{0x} =v_0cos(\alpha)[/tex]

Y axis:

[tex]Y= Y_0 +v_{y0} t - \frac{g}{2} t^2[/tex]

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, [tex]Y=0[/tex], then

[tex]0= Y_0 +v_{y0} t - \frac{g}{2} t^2[/tex]

[tex]0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2[/tex]

Caculate the segcond degree equation to obtain the two possible values for t:

[tex]t_1= 10.18 \\t_2= -0.04046[/tex]

But, in physics, time it could not be negative, so we take [tex]t_1= 10.18[/tex]

Caculate now:

[tex]X=79m/s*cos(\39)*10.18s= 624.996 m[/tex]

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

[tex]v_0= 79m/s+13m/s= 92m/s[/tex]

Using the same procedure that item A, caculate X

First, we need to know the new time

[tex]0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2[/tex]

And we obtain:

[tex]t_1=11.845s\\t_2=-0.041s[/tex]

One more time, we take the positive time: [tex]t_1=11.845s[/tex]

Finally:

[tex]X=92m/s *cos(39)*11.845s=846.887 m[/tex]