1 Answer

Felipecao · Answer 1 · 2019-03-19T20:09:16+0000

4. The Coyote has an initial position vector of
$\vec r_0=(15.5\,\mathrm m)\,\vec\jmath$ .

4a. The Coyote has an initial velocity vector of
$\vec v_0=\left(3.5\,(\mathrm m)/(\mathrm s)\right)\,\vec\imath$ . His position at time
is given by the vector

$\vec r=\vec r_0+\vec v_0t+\frac12\vec at^2$

where
$\vec a$ is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
$\vec a=-g\,\vec\jmath$ where
$g=9.80\,(\mathrm m)/(\mathrm s^2)$ . Splitting up the position vector into components, we have
$\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

$r_x=\left(3.5\,(\mathrm m)/(\mathrm s)\right)t$

$r_y=15.5\,\mathrm m-\frac g2t^2$

The Coyote hits the ground when
r_y=0 :

$15.5\,\mathrm m-\frac g2t^2=0\implies t=1.8\,\mathrm s$

4b. Here we evaluate
r_x at the time found in (4a).

$r_x=\left(3.5\,(\mathrm m)/(\mathrm s)\right)(1.8\,\mathrm s)=6.3\,\mathrm m$

5. The shell has initial position vector
$\vec r_0=(1.52\,\mathrm m)\,\vec\jmath$ , and we're told that after some time the bullet (now separated from the shell) has a position of
$\vec r=(3500\,\mathrm m)\,\vec\imath$ .