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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.

user694587
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  • Can you solve simpler problems like how long it takes a stone to fall to the ground after you throw it straight upward at $10\ \mathrm{ms}^{-1}$? You say you looked at other examples "to no avail", but that doesn't give any clue as to what is causing you difficulty. Whereas if you showed one of these examples and told us where you stopped understanding it, we'd have a clue about what kind of help would be useful to you. – David K Sep 03 '19 at 01:18

3 Answers3

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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.

user0102
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  • Slightly simpler is to use symmetry and solve for $y'(t)=-y'(0)$. – amd Sep 03 '19 at 01:50
  • Thanks for the input. Iv been playing around with your suggestions but am still unsure what values im plugging in and where im plugging them in. I assume we are using either 10 or 9.8 for our gravity value. – user694587 Sep 05 '19 at 23:19
  • In the present case, we have $h_{0} = h_{1} = 0m$, $v_{0} = v_{1} = 20m/s$, $\theta_{0} = \pi/3$ and $\theta_{1} = \pi/6$. You may assume that $g = 10m/s^{2}$. – user0102 Sep 05 '19 at 23:26
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(a) Using standard kinematic formula for vertical motion gives $$t=\frac {2v\sin\theta}g$$ where $t$ is total flight time. Since $v$ is the same, the stone launched at $30^\circ$ takes a shorter time.

(b) The range for the projectile is given by $$R=\frac {v^2\sin2\theta}g$$ which can be easily proven.

Since $\sin 2(60^\circ)=\sin 2(30^\circ)$, and $v$ is the same in both cases, both projectiles land at the same point.

Hypergeometricx
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The equation of vertical motion is

$$y=v\sin\theta\,t-g\frac{t^2}2$$ so that the flight time is

$$t=\frac{2v\sin\theta}g.$$

The faster is with the smallest angle.

The horizontal motion equation gives

$$x=v\cos\theta\,t=\frac{2v^2\sin\theta\cos\theta}g=\frac{v^2\sin2\theta}g.$$

Indeed, $\sin60°=\sin120°$.