A stone is thrown with a velocity of 20 ms^(-1) and at an angle of 60°. A second stone is thrown at the same time and place,with the same velocity but an angle of 30°.
(a) Find which stone finishes its flight first and by how long.
(b) Show that both stones land at the same place.
I'm having difficulty with projectile motion. Iv looked at numerous examples but to no avail. I need some help with where to start and some working for this question.
To solve this kind of problem, you have to consider the two components of the displacement. Precisely, fixing ''the most convenient coordinate system'', we obtain the following equations for the movement of the first stone \begin{align*} \begin{cases} x_{0}(t) = v_{0}\cos(\theta_{0})t\\ y_{0}(t) = h_{0} + v_{0}\sin(\theta_{0})t - \displaystyle \frac{gt^{2}}{2} \end{cases} \end{align*} Analogously, we obtain the following equations for the movement of the second stone \begin{align*} \begin{cases} x_{1}(t) = v_{1}\cos(\theta_{1})t\\ y_{1}(t) = h_{1} + v_{1}\sin(\theta_{1})t - \displaystyle \frac{gt^{2}}{2} \end{cases} \end{align*}
Where $h_{k}$ indicates the initial height, $v_{k}$ indicates the initial speed, $g$ indicates the gravity and $t$ stands for the time. Now plug in the given values and solve the equations $y_{k} = 0$ in order to obtain the traveling time for each stone. Once you have each $t_{k}$, you may use it to obtain the displacement of each stone on the $x$-axis.