Solved Problem on Two-dimensional Motion
advertisement   



Three identical spheres are launched from the same height H with the same speeds. Sphere A is thrown vertically downward, B is thrown vertically upward, and C is thrown horizontally. Which one gets to the ground at a higher speed, neglecting the resistance of the air.


Problem data:
  • Initial speed:    v0;
  • Height of the release point:    h.
Solution
  • Sphere A:
We choose a reference frame pointing upward with the origin on the ground, the acceleration due to gravity, and the velocity of the sphere, which are negative, in the opposite direction of the reference frame (Figure 1). The motion of the sphere is a vertical launch downward under the action of acceleration due to gravity, applying the equation of velocity as a function of displacement, its final speed will be
\[ \begin {gather} \bbox[#99CCFF,10px] {v^2=v_0^2+2a\Delta S} \tag{I} \end{gather} \]
\[ \begin{gather} v^2=(-v_0)^2+2(-g)\Delta S\\[5pt] v^2=v_0^2-2g(0-h)\\[5pt] v^2=v_0^2-2g(-h)\\[5pt] v^2=v_0^2+2gh\\[5pt] v=\sqrt{v_0^2+2gh} \tag{II} \end{gather} \]
Figure 1
  • Sphere B:
The motion of the sphere is divided into two parts. The first part is a vertical upward launch, the sphere rises to a height of H until its speed equals zero. The second part of the motion is a free fall from the rest.
In the first part, the initial speed of the launch is positive, it is oriented in the same direction as the reference frame, and the final speed is zero, vH = 0, which is the instant the sphere stops and reverses the motion to begin to fall (Figure 2-A).

Figure 2

Writing the expression (I) for the first part, we find the height hit by the sphere
\[ \begin{gather} v^2=v_0^2+2a\Delta S\\[5pt] v_H^2=v_0^2+2(-g)\Delta S\\[5pt] 0^2=v_0^2-2g(H-h)\\[5pt] 0=v_0^2-2gH+2gh\\[5pt] 2gH=v_0^2+2gh\\[5pt] H=\frac{v_0^2+2gh}{2g} \tag{III} \end{gather} \]
In the second part, free fall, the initial speed will be the final speed of the first part, equal to zero, v0H = vH = 0, and the initial height will be given by the expression (III). Applying the expression (I), we have the speed with which the sphere reaches the ground
\[ \begin{gather} v^2=v_0^2+2a\Delta S\\[5pt] v^2=v_{0H}^2+2(-g)\Delta S\\[5pt] v^2=0^2-2g(0-H)\\[5pt] v^2=0-2g(-H)\\[5pt] v^2=0+2gH\\[5pt] v^2=\cancel{2g}\left(\frac{v_0^2+2gh}{\cancel{2g}}\right)\\[5pt] v=\sqrt{v_0^2+2gh} \tag{IV} \end{gather} \]
  • Sphere C:
The motion should be decomposed in the x and y directions (Figure 3-A).

Figure 3

In the x direction, there is no acceleration acting on the sphere, it moves uniformly, for each time interval the displacements along the x-axis are the same (Δ x1=Δ x2=Δ x3=Δ x4=Δ x5 - Figura 3-B). The component of speed in that direction will be the speed of launching the sphere v0
\[ \begin{gather} v_x=v_0 \tag{V} \end{gather} \]
In the y direction, we have the acceleration due to gravity acting on the sphere, for the same time interval the displacements are increasing (Δ y1<Δ y2<Δ y3<Δ y4<Δ y5). Initially, the speed is zero v0y = 0, the speed in that direction is given by the expression (I), with a = −g
\[ \begin{gather} v_y^2=v_{0y}^2+2(-g)\Delta S_y\\[5pt] v_y^2=0^2-2g(0-h)\\[5pt] v_y^2=0-2g(-h)\\[5pt] v_y^2=2gh \tag{VI} \end{gather} \]
The speed with which the sphere reaches the ground is given by the sum of the components in the x and y directions given by the expressions (V) and (VI) respectively, using the Pythagorean Theorem (Figure 4)
\[ \begin{gather} v^2=v_{x}^2+v_y^2\\[5pt] v^2=v_0^2+2gh\\[5pt] v=\sqrt{v_0^2+2gh} \tag{VII} \end{gather} \]
Figure 4

Comparing expressions (II), (IV), and (VII), we see that all spheres come to the ground at the same speed.
advertisement   

Creative Commons License
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .