Solved Problem on Two-dimensional Motion
ES
FR
PT-BR
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From two points A and B, situated at a distance of 1000 m from each other, on the same
horizontal plane, two rockets are launched simultaneously. One departs from point B with an
initial velocity of 200 m/s directed upwards, and the other from point A in the direction of the
vertical passing through B, making an angle of 60° with the horizon. Determine:
a) The initial speed of rocket A so that it intercepts the second;
b) After how long does the meeting of the two rockets take place;
c) At what height does the meeting take place;
d) Check if this encounter takes place during the ascent or fall of the first rocket.
a) The initial speed of rocket A so that it intercepts the second;
b) After how long does the meeting of the two rockets take place;
c) At what height does the meeting take place;
d) Check if this encounter takes place during the ascent or fall of the first rocket.
Problem data:
- Distance between launch points A and B de lançamento: d = 1000 m;
- Launch angle of rocket A: θ = 60°;
- Initial speed of rocket B: v0B = 200 m/s,
- Acceleration due to gravity: g = 9.8 m/s2.
We choose a frame of reference on the ground with the axis Ox pointing to the right and Oy pointing upwards, the acceleration due to gravity pointing downwards, and the point where the rocket A is launched is at (x0A, y0A) = (0, 0), and the rocket B is at (x0B, y0B) = (0, 1000), (Figure 1).
Solution
The motion of the rocket launched from A can be decomposed in the x and y
directions. The initial velocity v0A, with which it is launched, has components
in the x and y directions (Figure 2)
\[
\begin{gather}
v_{0Ax}=v_{0A}\cos 60°\\[10pt]
v_{0Ay}=v_{0A}\sin 60°
\end{gather}
\]
From the Trigonometry,
\( \cos 60°=\dfrac{1}{2} \)
and
\( \sin 60°=\dfrac{\sqrt{3\;}}{2} \)
\[
\begin{gather}
v_{0Ax}=\frac{1}{2}v_{0A} \tag{I}
\end{gather}
\]
\[
\begin{gather}
v_{0Ay}=\frac{\sqrt{3\;}}{2}v_{0A} \tag{II}
\end{gather}
\]
In the x direction, there is no acceleration acting on the rocket, it is in uniform motion and its
motion is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S=S_{0}+vt}
\end{gather}
\]
as in uniform motion vAx = v0Ax is constant, we can substitute
vAx by the value of (I) and S0Ax = 0
\[
\begin{gather}
S_{Ax}=S_{0Ax}+v_{Ax}t\\[5pt]
S_{Ax}=0+\frac{1}{2}v_{0A}t\\[5pt]
S_{Ax}=\frac{1}{2}v_{0A}t \tag{III}
\end{gather}
\]
In the y direction, the rocket is under the action of acceleration due to gravity, so the
equations of displacement as a function of time and velocity as a function of time with constant
acceleration are given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S=S_{0}+v_{0}t-\frac{g}{2}t^2} \tag{IV}
\end{gather}
\]
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=v_{0}-gt}
\end{gather}
\]
substituting v0Ay with the value given in (II) and
S0Ay = 0
\[
\begin{gather}
S_{Ay}=S_{0Ay}+v_{0Ay}t-\frac{g}{2}t^2\\[5pt]
S_{Ay}=0+\frac{\sqrt{3\;}}{2}v_{0A}t-\frac{\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{2}t^2\\[5pt]
S_{Ay}=\frac{\sqrt{3\;}}{2}v_{0A}t-\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2 \tag{V}
\end{gather}
\]
\[
\begin{gather}
v_{Ay}=v_{0ay}-gt\\[5pt]
v_{Ay}=\frac{\sqrt{3\;}}{2}v_{0A}-\left(9.8\;\mathrm{\small{\frac{m}{s^2}}}\right)t \tag{VI}
\end{gather}
\]
the minus sign indicates that the acceleration due to gravity is in the opposite direction of the reference
frame).
In Figure 3, we see that in the movement along the x direction we have that for equal time
intervals, we have equal displacement
(Δx1 = Δx2 =
Δx3 = Δx4 = Δx5 =
Δx6).
In the y direction, at the instant the rocket is launched, we have the velocity
vy starts to decrease. for equal time intervals, we have smaller displacements
(Δy1 > Δy2 > Δy3 >
Δy4 > Δy5 > Δy6).
The rocket launched from B only has motion along the y-axis, it is under the action of acceleration due to gravity in Uniformly Accelerated Motion, applying equation (IV).
The rocket launched from B only has motion along the y-axis, it is under the action of acceleration due to gravity in Uniformly Accelerated Motion, applying equation (IV).
\[
\begin{gather}
S_B=S_{0B}+v_{0B}t-\frac{g}{2}t^2
\end{gather}
\]
substituting v0B with the value given in the problem and S0B = 0
\[
\begin{gather}
S_B=0+\left(200\;\mathrm{\small{\frac{m}{s}}}\right)t-\frac{\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{2}t^2\\[5pt]
S_B=\left(200\;\mathrm{\small{\frac{m}{s}}}\right)t-\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2 \tag{VII}
\end{gather}
\]
a) For the meeting to occur, we have the condition
\[
\begin{gather}
S_{Ay}=S_B
\end{gather}
\]
\[
\begin{gather}
\frac{\sqrt{3\;}}{2}v_{0A}t-\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2=\left(200\;\mathrm{\small{\frac{m}{s}}}\right)t-\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2\\[5pt]
\frac{\sqrt{3\;}}{2}v_{0A}t=\left(200\;\mathrm{\small{\frac{m}{s}}}\right)t-\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2+\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2\\[5pt]
\frac{\sqrt{3\;}}{2}v_{0A}\cancel t=\left(200\;\mathrm{\small{\frac{m}{s}}}\right)\cancel t\\[5pt]
v_{0A}=\frac{2\times 200\;\mathrm{\frac{m}{s}}}{\sqrt{3\;}}\\[5pt]
v_{0A}=\frac{400\;\mathrm{\frac{m}{s}}}{\sqrt{3\;}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_{0A}\approx 231\;\mathrm{m/s}}
\end{gather}
\]
b) As the rocket departing from B rises vertically, the rocket departing from A must travel a distance of 1000 m in the x direction to intercept it, substituting the value of the previous item and SAx = 1000 m in equation (III)
\[
\begin{gather}
1000\;\mathrm m=\frac{1}{2}\times\left(231\;\mathrm{\small{\frac{m}{s}}}\right)t\\[5pt]
t=\frac{2\times 1000\;\mathrm{\cancel m}}{231\;\mathrm{\frac{\cancel m}{s}}}\\[5pt]
t=\frac{2000}{231}\;\mathrm s
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{t\approx 8.6\;\mathrm{s}}
\end{gather}
\]
c) Rocket B will climb vertically until the encounter occurs, substituting the value of the previous item in equation (VII)
\[
\begin{gather}
S_B=200\;\mathrm{\small{\frac{m}{\cancel s}}}\times 8.6\;\mathrm{\cancel s}-4.9\;\mathrm{\small{\frac{m}{s^2}}}\times(8.6\;\mathrm s)^2\\[5pt]
S_B=1720\;\mathrm m-4.9\;\mathrm{\small{\frac{m}{\cancel{s^2}}}}\times 73.9\;\mathrm{\cancel{s^2}}\\[5pt]
S_B=1720-362
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{S_B\approx 1358\;\mathrm m}
\end{gather}
\]
Note: We could find the same value by substituting the velocity of item (a) and the
time interval found in (b) into equation (V).
d) If the instant of time of the meeting is smaller than the time interval for rocket A to reach the maximum height, the meeting will occur during the ascent, if the instant of time is greater, the meeting will occur during the descent. When the rocket starting from A reaches its maximum height, the component of its velocity in the y direction is equal to zero, vAy = 0, the time it takes for the rocket from A to reach that height will be obtained by substituting this condition and the speed of part (a) in equation (VI)
\[
\begin{gather}
0=\frac{\sqrt{3\;}}{2}\times\left(231\;\mathrm{\small{\frac{m}{s}}}\right)-\left(9.8\;\mathrm{\small{\frac{m}{s^2}}}\right)t\\[5pt]
\frac{\sqrt{3\;}}{2}\times\left(231\;\mathrm{\small{\frac{m}{s}}}\right)=\left(9.8\;\mathrm{\small{\frac{m}{s^2}}}\right)t\\[5pt]
t=\frac{\sqrt{3\;}\times\left(231\;\mathrm{\frac{\cancel m}{\cancel s}}\right)}{2\times\left(9.8\;\mathrm{\frac{\cancel m}{s^{\cancel 2}}}\right)}\\[5pt]
t\approx 20.4\;\mathrm s
\end{gather}
\]
As the time interval for the rocket to reach maximum height is longer than the time interval for the
meeting to occur, the meeting occurs during the
ascent of the first rocket.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .