Answer:
A.) The arrow`s range is 624,996 m
B.) The arrow`s range is 846.887 m, when the horse is galloping
Explanation:
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.
<u>Equations</u>
X axis:
![X=v_{ox}*t](https://tex.z-dn.net/?f=X%3Dv_%7Box%7D%2At)
![v_{0x} =v_0cos(\alpha)](https://tex.z-dn.net/?f=v_%7B0x%7D%20%3Dv_0cos%28%5Calpha%29)
Y axis:
![Y= Y_0 +v_{y0} t - \frac{g}{2} t^2](https://tex.z-dn.net/?f=Y%3D%20Y_0%20%2Bv_%7By0%7D%20t%20-%20%5Cfrac%7Bg%7D%7B2%7D%20t%5E2)
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,
, then
![0= Y_0 +v_{y0} t - \frac{g}{2} t^2](https://tex.z-dn.net/?f=0%3D%20Y_0%20%2Bv_%7By0%7D%20t%20-%20%5Cfrac%7Bg%7D%7B2%7D%20t%5E2)
![0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2](https://tex.z-dn.net/?f=0%3D2.4m%2B79m%2Fs%2Asin%2839%29t-%281%2F2%2A9.81m%2Fs%5E2%29t%5E2)
Caculate the segcond degree equation to obtain the two possible values for t:
![t_1= 10.18 \\t_2= -0.04046](https://tex.z-dn.net/?f=t_1%3D%2010.18%20%5C%5Ct_2%3D%20-0.04046)
But, in physics, time it could not be negative, so we take ![t_1= 10.18](https://tex.z-dn.net/?f=t_1%3D%2010.18)
Caculate now:
![X=79m/s*cos(\39)*10.18s= 624.996 m](https://tex.z-dn.net/?f=X%3D79m%2Fs%2Acos%28%5C39%29%2A10.18s%3D%20624.996%20m)
B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.
![v_0= 79m/s+13m/s= 92m/s](https://tex.z-dn.net/?f=v_0%3D%2079m%2Fs%2B13m%2Fs%3D%2092m%2Fs)
Using the same procedure that item A, caculate X
First, we need to know the new time
![0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2](https://tex.z-dn.net/?f=0%3D2.4m%2B92m%2Fs%2Asin%2839%29t-%281%2F2%2A9.81m%2Fs%5E2%29t%5E2)
And we obtain:
![t_1=11.845s\\t_2=-0.041s](https://tex.z-dn.net/?f=t_1%3D11.845s%5C%5Ct_2%3D-0.041s)
One more time, we take the positive time: ![t_1=11.845s](https://tex.z-dn.net/?f=t_1%3D11.845s)
Finally:
![X=92m/s *cos(39)*11.845s=846.887 m](https://tex.z-dn.net/?f=X%3D92m%2Fs%20%2Acos%2839%29%2A11.845s%3D846.887%20m)