Question

Can anyone solve these for my by using unit vectors? Can you also please show your work

Answer 1

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 $\vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath$. His position at time $t$ is given by the vector

$\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2$

where $\vec a$ is the Coyote's acceleration vector at time $t$. He experiences acceleration only in the downward direction because of gravity, and in particular $\vec a=-g\,\vec\jmath$ where $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$. Splitting up the position vector into components, we have $\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t$

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

The Coyote hits the ground when $r_y=0$:

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

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

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m$

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 $\vec r=(3500\,\mathrm m)\,\vec\imath$.

5a. The vertical component of the shell's position vector is

$r_y=1.52\,\mathrm m-\dfrac g2t^2$

We find the shell hits the ground at

$1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s$

5b. The horizontal component of the bullet's position vector is

$r_x=v_0t$

where $v_0$ 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 $v_0$:

$3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}$