4. The Coyote has an initial position vector of [tex]\vec r_0=(15.5\,\mathrm m)\,\vec\jmath[/tex].
4a. The Coyote has an initial velocity vector of [tex]\vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath[/tex]. His position at time [tex]t[/tex] is given by the vector
[tex]\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2[/tex]
where [tex]\vec a[/tex] is the Coyote's acceleration vector at time [tex]t[/tex]. He experiences acceleration only in the downward direction because of gravity, and in particular [tex]\vec a=-g\,\vec\jmath[/tex] where [tex]g=9.80\,\frac{\mathrm m}{\mathrm s^2}[/tex]. Splitting up the position vector into components, we have [tex]\vec r=r_x\,\vec\imath+r_y\,\vec\jmath[/tex] with
[tex]r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t[/tex]
[tex]r_y=15.5\,\mathrm m-\dfrac g2t^2[/tex]
The Coyote hits the ground when [tex]r_y=0[/tex]:
[tex]15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s[/tex]
4b. Here we evaluate [tex]r_x[/tex] at the time found in (4a).
[tex]r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m[/tex]
5. The shell has initial position vector [tex]\vec r_0=(1.52\,\mathrm m)\,\vec\jmath[/tex], and we're told that after some time the bullet (now separated from the shell) has a position of [tex]\vec r=(3500\,\mathrm m)\,\vec\imath[/tex].
5a. The vertical component of the shell's position vector is
[tex]r_y=1.52\,\mathrm m-\dfrac g2t^2[/tex]
We find the shell hits the ground at
[tex]1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s[/tex]
5b. The horizontal component of the bullet's position vector is
[tex]r_x=v_0t[/tex]
where [tex]v_0[/tex] is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for [tex]v_0[/tex]:
[tex]3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}[/tex]
