Answer:
[tex]\Delta x=22.67786838m[/tex]
Explanation:
Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:
[tex]y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2[/tex] (1)
Where:
[tex]y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis[/tex]
[tex]y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis[/tex]
[tex]v_o=initial\hspace{3}velocity[/tex]
[tex]t=travel\hspace{3}time[/tex]
[tex]g=gravity\hspace{3}constant[/tex]
[tex]\theta=Initial\hspace{3}launch\hspace{3}angle[/tex]
In this case:
[tex]\theta=0[/tex]
Because the dog jumps horizontally
Let's asume the gravity constant as:
[tex]g=9.8[/tex]
[tex]y=0[/tex]
Because when the dog reach the base the height is 0
[tex]y_o=70[/tex]
[tex]v_o=6[/tex]
Now let's replace the data in (1)
[tex]y_o-\frac{1}{2} *(9.8)*t^2+70[/tex]
Isolating t:
[tex]t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473[/tex]
Finally let's find the horizontal displacement using this equation:
[tex]\Delta x=v_o*cos(\theta)*t[/tex]
Replacing the data:
[tex]\Delta x=6*1*3.77964473=22.67786838m[/tex]
