Velocity and Acceleration — Question 2

Question 2 – Solution

Strategy. Integrate constant acceleration twice and apply the initial data; landing occurs when the vertical coordinate is zero.

Step 1: Motion functions $$\mathbf v(t)=\left\langle 30,40,50-32t\right\rangle.$$ Integrating again and using $\mathbf r(0)=\left\langle 0,0,5\right\rangle$, $$\boxed{\mathbf r(t)=\left\langle 30t,40t,5+50t-16t^2\right\rangle}.$$

Step 2: Maximum height The vertical velocity is zero when $$50-32t=0\quad\Longrightarrow\quad t=\frac{25}{16}.$$ Then $$z=5+50\left(\frac{25}{16}\right)-16\left(\frac{25}{16}\right)^2 =\boxed{\frac{705}{16}\text{ ft}}.$$

Step 3: Landing Set $z=0$: $$5+50t-16t^2=0 \quad\Longrightarrow\quad t=\frac{25\pm\sqrt{705}}{16}.$$ Only the positive root is physical: $$\boxed{t_L=\frac{25+\sqrt{705}}{16}}.$$ Thus the horizontal landing point is $$\boxed{(x,y)=\left(30t_L,40t_L\right)}.$$ Substitution into the vertical equation verifies height zero.