Answer
a. $x=(100cos40^\circ) t$, $y=6+(100sin40^\circ) t-16t^2$
b. $t=1sec$, $x \approx76.6\ ft$, $y \approx54.3\ ft$;
$t=2sec$, $x \approx153.2\ ft$, $y \approx70.6\ ft$;
$t=3sec$, $x \approx229.8\ ft$, $y \approx54.8\ ft$.
c. $t\approx4.1\ sec$, $x \approx314.1\ ft$
d. $t=2.0\ sec$, $y\approx70.6\ ft$
Work Step by Step
a. Given $v_0=100, \theta=40^\circ, h=6$ and using the formula in the exercise, we have $x=(100cos40^\circ) t$ and $y=6+(100sin40^\circ) t-16t^2$
b. At $t=1sec$, we have $x=(100cos40^\circ)(1)\approx76.6\ ft$ and $y=6+(100sin40^\circ)(1)-16(1)^2\approx54.3\ ft$.
At $t=2sec$, we have $x=(100cos40^\circ)(2)\approx153.2\ ft$ and $y=6+(100sin40^\circ)(2)-16(2)^2\approx70.6\ ft$.
At $t=3sec$, we have $x=(100cos40^\circ)(3)\approx229.8\ ft$ and $y=6+(100sin40^\circ)(3)-16(3)^2\approx54.8\ ft$.
c. To find the time of flight, let $y=0$. We have $6+(100sin40^\circ) t-16t^2=0$. Solve this equation using a formula or graphically as shown in the top figure. Choosing the positive solution, we have $t\approx4.1\ sec$ and $x=(100cos40^\circ)(4.1)\approx314.1\ ft$
d. Graph the equations as shown in the bottom figure. We can find that the maximum happens at $t=2.0\ sec$, giving $x\approx153.2$ and $y\approx70.6\ ft$ as the maximum height.
