Solved Problem on Two-dimensional Motion
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A basketball player throws the ball toward the hoop at a distance of 4.6 m making a 60° angle with the horizontal. The basket is at a height of 3.05 m and the ball is 2.25 m from the ground when it leaves the player's hands. Calculate the initial speed of the ball and the time spent by the ball to go from the player's hands to the hoop.


Problem Data:
  • Distance from player to hoop:    D = 4.6 m;
  • Height of the ball to the ground:    h = 2.25 m;
  • Height of the hoop to the ground:    H = 3.05 m;
  • Angle of the ball launching:    θ = 60°;
  • Acceleration due to gravity:    g = 9.8 m/s2.
Problem diagram:

We choose a frame of reference from the point where the ball is thrown, with the Ox axis pointing to the right and the Oy axis pointing upward; the acceleration due to gravity is directed downward. The ball is thrown from the origin of the reference frame, (x0, y0) = (0, 0). The y1 coordinate of the point where the basket is located is the difference between the height of the basket and the height of the player's hands, y1 = H−h = 3,05−2,25 = 0,8.

Figure 1

Solution

The initial velocity v0 can be decomposed in the x and y directions
\[ \begin{gather} v_{0x}=v_0\cos 60°\\[10pt] v_{0y}=v_0\sin 60° \end{gather} \]
Figure 2

From the trigonometry,   \( \cos 60°=\dfrac{1}{2} \)   and   \( \operatorname{sen}60°=\dfrac{\sqrt{3\;}}{2} \)
\[ \begin{gather} v_{0x}=\frac{1}{2}v_0 \tag{I} \end{gather} \]
\[ \begin{gather} v_{0y}=\frac{\sqrt{3\;}}{2}v_0 \tag{II} \end{gather} \]
In the x direction, there is no acceleration acting on the ball, it is in the Uniform Rectilinear Motion given by the equation
\[ \begin{gather} \bbox[#99CCFF,10px] {S_x=S_{0x}+v_xt} \end{gather} \]
since in the motion, vx = v0x is constant, we can substitute vx with the value of (I) and S0x = 0
\[ \begin{gather} S_x=0+\frac{1}{2}v_0t\\[5pt] S_x=\frac{1}{2}v_0t \tag{III} \end{gather} \]
In the y direction, the ball is under the action of acceleration due to gravity, it is in free-fall given by the equation
\[ \begin{gather} \bbox[#99CCFF,10px] {S_y=S_{0y}+v_{0y}t-\frac{g}{2}t^{2}} \end{gather} \]
with constant −g (the negative sign indicates that the acceleration due to gravity is in the opposite direction of the frame of reference), substituting v0y with the value given in (II) and S0y = 0
\[ \begin{gather} S_y=0+\frac{\sqrt{3\;}}{2}v_0t-\frac{(9.8\;\mathrm{m/s²})}{2} t^2\\[5pt] S_y=\frac{\sqrt{3\;}}{2}v_0t-\left(4.9\;\mathrm{\frac{m}{s^2}}\right)t^2 \tag{IV} \end{gather} \]
From Figure 3, we see that in the motion along the x direction, we ​​have to for intervals of equal times we have equal displacement intervals (Δx1 = Δx2 = Δx3 = Δx4 = Δx5 = Δx6 = Δx7). In the y direction, we have that during the rise for equal time intervals, we have lower displacements, the ball is being braked by the action of gravity (Δy1 > Δy2 > Δy3 > Δy4) until the vy velocity becomes zero. Then the action of gravity begins to pull the ball back toward the hoop with accelerated speed, so for an equal interval of time we have increasing displacement intervals (Δy5 < Δy6 < Δy7).
Figure 3

Substituting the distance from the player to the basket, Sx = D = 4.6 m, into equation (III)
\[ \begin{gather} 4.6\;\mathrm m=\frac{1}{2}v_0t \tag{V} \end{gather} \]
Substituting the height of the hoop, final height, Sx = H− = 0.8 m, into equation (IV)
\[ \begin{gather} 0.8\;\mathrm m=\frac{\sqrt{3\;}}{2}v_0t-\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2 \tag{VI} \end{gather} \]
The equations (V) and (VI) can be written as a system of two equations to two unknowns (v0 and t)
\[ \begin{gather} \left\{ \begin{array}{l} \dfrac{1}{2}v_0t=4,6\;\mathrm m\\ -\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2+\dfrac{\sqrt{3\;}}{2}v_0t=0.8\;\mathrm m \end{array} \right. \end{gather} \]
solving the first equation of the system for v0
\[ \begin{gather} v_0=\frac{2\times 4.6\;\mathrm m}{t} \tag{VII} \end{gather} \]
and substituting the equation (VII) into the second system equation
\[ \begin{gather} -\left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2+\frac{\sqrt{3\;}}{\cancel 2}\times\frac{\cancel 2\times 4.6\;\mathrm m}{\cancel t}\cancel t=0.8\;\mathrm m\\[5pt] \left(4.9\;\mathrm{\small{\frac{m}{s^2}}}\right)t^2=1,7\times 4.6\;\mathrm m-0.8\;\mathrm m\\[5pt] t^2=\frac{7\;\mathrm{\cancel m}} {4.9\;\mathrm{\small{\frac{\cancel m}{s^2}}}}\\[5pt] t^2=1.4\;\mathrm{s^2}\\[5pt] t=\sqrt{1.4\;\mathrm{s^2}\;} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {t\approx 1.2\;\mathrm s} \end{gather} \]
substituting this value into the equation (VII)
\[ \begin{gather} v_0=\frac{2\times 4.6\;\mathrm m}{1.2\;\mathrm s} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v_0\approx 7.7\;\mathrm{m/s}} \end{gather} \]
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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 .