Solved Problem on Two-dimensional Motion
ES
FR
PT-BR
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A ball rolls on the roof of a house until it falls over the edge with speed v0. The
height of the point from which the ball falls is equal to H, and the angle of inclination of the
roof with the vertical is equal to θ. Calculate:
a) The time required for the ball to hit the ground;
b) The horizontal distance from the house, where the ball hits the ground;
c) The equation of motion;
d) The speed with which the ball hits the ground.
a) The time required for the ball to hit the ground;
b) The horizontal distance from the house, where the ball hits the ground;
c) The equation of motion;
d) The speed with which the ball hits the ground.
Problem data:
- Initial speed of the ball: v0;
- Roof edge height: H;
- Roof inclination angle: θ.
We choose a frame of reference with origin on the floor with the axis Ox pointing to the right and Oy pointing upwards, the acceleration due to gravity pointing downwards, and the point where the ball falls from the roof at (x0, y0) = (0, H), (Figure 1).
Solution
The motion can be decomposed in the x and y directions. The initial velocity
v0, with which the ball rolls off the roof, has components in the x and
y directions (Figure 2)
\[
\begin{gather}
v_{0x}=v_0\sin\theta \tag{I}\\[10pt]
v_{0y}=-v_0\cos\theta \tag{II}
\end{gather}
\]
where the component in x is proportional to sine, and in y to cosine, contrary to what is
usually done because the angle θ was measured relative to the y-axis.
In the x direction, there is no acceleration acting on the ball, it is in Uniform Rectilinear Motion given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S_x=S_{0x}+v_xt}
\end{gather}
\]
since in uniform motion vx = v0x is constant, we can substitute
vx by the value of (I) and S0x = 0
\[
\begin{gather}
S_x=0+v_0\sin\theta t\\[5pt]
S_x=v_0\sin\theta t\tag{III}
\end{gather}
\]
In the y direction, the ball is under the action of the acceleration due to gravity, so it is in free
fall given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S_y=S_{0y}+v_{0y}t-\frac{g}{2}t^2}
\end{gather}
\]
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v_y=v_{0y}-gt}
\end{gather}
\]
substituting v0y with the value given in (II) and
S0y = H
\[
\begin{gather}
S_y=H+(-v_0\cos\theta)t-\frac{g}{2}t^2\\[5pt]
S_y=H-v_0\cos\theta t-\frac{g}{2}t^2\tag{IV}
\end{gather}
\]
\[
\begin{gather}
v_y=-v_0\cos\theta-gt\tag{V}
\end{gather}
\]
with constant −g (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). In the y direction at the instant the
ball falls from the roof, we have the velocity vy starts to increase from zero
under the action of gravity, for equal time intervals, we have increasingly larger displacements
(Δy1 < Δy2 < Δy3 <
Δy4).
a) The time interval for the ball to hit the ground will be obtained from equation (IV) with the condition that the height on the ground is zero, Sy = 0
\[
\begin{gather}
0=H-v_0\cos\theta t-\frac{g}{2}t^2
\end{gather}
\]
this is a Quadratic Equation where the unknown is the value of t.
Solution of \( -\dfrac{g}{2}t^2-v_0\cos\theta t+H=0 \)
\[
\begin{gather}
\Delta =b^2-4ac=(-v_0\cos\theta)^2-\cancelto{2}{4}\left(-{\frac{g}{\cancel 2}}\right)H=v_0^2\cos^2\theta+2gH
\end{gather}
\]
\[
\begin{split}
t=\frac{-b\pm \sqrt{\Delta \;}}{2a} &=\frac{-(-v_0\cos\theta)\pm \sqrt{v_0^2\cos^2\theta+2gH}}{\cancel{2}\left(-{\dfrac{g}{\cancel{2}}}\right)}=\\
&=\frac{-v_0\cos\theta\pm\sqrt{v_0^2\cos^2\theta+2gH}}{g}
\end{split}
\]
the roots are
\[
\begin{gather}
t_1=\frac{-v_0\cos\theta+\sqrt{v_0^2\cos^2\theta+2gH}}{g}
\\[5pt]\text{or}\\[5pt]
t_2=\frac{-v_0\cos\theta-\sqrt{v_0^2\cos^2\theta+2gH}}{g}
\end{gather}
\]
neglecting the second root, which has a negative value, t2 < 0, the time for the ball to hit the ground will be
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{t=\frac{-v_0\cos\theta+\sqrt{v_0^2\cos^2\theta+2gH}}{g}}
\end{gather}
\]
b) The time interval calculated above, for the ball to fall to the ground, is also the time interval it will take to go from the origin to point D along the x-axis, substituting the answer to the previous item in the expression (III)
\[
\begin{gather}
D=v_0\sin\theta\left(\frac{-v_0\cos\theta+\sqrt{v_0^2\cos^2\theta+2gH}}{g}\right)\\[5pt]
D=v_0\left(\frac{-v_0\cos\theta\sin\theta+\sin\theta\sqrt{v_0^2\cos^2\theta+2gH}}{g}\right)\\[5pt]
D=v_0\left(\frac{-v_0\cos\theta\sin\theta+\sqrt{\sin^2\theta(v_0^2\cos^2\theta+2gH)}}{g}\right)\\[5pt]
D=v_0\left(\frac{-v_0\cos\theta\sin\theta+\sqrt{v_0^2\cos^2\theta\sin^2\theta+2gH\sin^2\theta}}{g}\right)\tag{VI}
\end{gather}
\]
From the Trigonometric identities
\[
\begin{gather}
\sin(\theta+\theta)=\sin\theta\cos\theta+\sin\theta\cos\theta\\[5pt]
\sin(2\theta)=2\sin\theta\cos\theta\\[5pt]
\cos\theta\sin\theta=\frac{\sin 2\theta}{2}\tag{VII}
\end{gather}
\]
squaring the expression (VII) on both sides of the equality
\[
\begin{gather}
(\cos\theta\sin\theta)^2=\left(\frac{\sin 2\theta}{2}\right)^2\\[5pt]
\cos^2\theta\sin^2\theta=\frac{\sin^2\theta}{4} \tag{VIII}
\end{gather}
\]
Substituting the expressions (VII) and (VIII) into equation (VI)
\[
\begin{gather}
D=v_0\left(\frac{-v_0\dfrac{\sin 2\theta}{2}+\sqrt{v_0^2\dfrac{\sin^2 2\theta}{4}+2gH\sin^2\theta}}{g}\right)\\[5pt]
D=v_0\left(\frac{\dfrac{-v_0\sin 2\theta}{2}+\sqrt{\dfrac{v_0^2\sin^2 2\theta+8gH\sin^2\theta}{4}}}{g}\right)\\[5pt]
D=v_0\left(\frac{\dfrac{-v_0\sin 2\theta}{2}+\dfrac{1}{2}\sqrt{v_0^2\sin^2 2\theta+8gH\sin^2\theta}}{g}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{D=v_0\left(\frac{-v_0\sin 2\theta+\sqrt{v_0^2\sin^2 2\theta+8gH\sin^2\theta}}{2g}\right)}
\end{gather}
\]
c) To obtain the equation of motion (Figure 1), we have y as a function of x, or y = f(x), using equations (III) and (IV) for the motions in x and y.
\[
\left\{
\begin{array}{l}
S_x=v_0\sin\theta t\\
S_y=H-v_0\cos\theta t-\dfrac{g}{2}t^2
\end{array}
\right.
\]
solving the first equation for time t
\[
\begin{gather}
t=\frac{S_x}{v_0\sin\theta}
\end{gather}
\]
substituting this value in the second equation of the system
\[
\begin{gather}
S_y=H-v_0\cos\theta\frac{S_x}{v_0\sin\theta}-\frac{g}{2}\left(\frac{S_x}{v_0\sin\theta}\right)^2\\[5pt]
S_y=H-\frac{\cos\theta}{\sin\theta}S_x-\frac{g}{2v_0^2\sin^2\theta}S_x^2
\end{gather}
\]
From the Trigonometry
\[
\begin{gather}
\tan\theta=\dfrac{\sin\theta}{\cos\theta}\Rightarrow \dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{S_y=H-\frac{1}{\tan\theta}S_x-\frac{g}{2v_0^2\sin^2\theta}S_x^2}
\end{gather}
\]
Making the association with a Quadratic Function of the type
\( y=ax^{2}+bx+c \)
\[
\begin{array}{c}
S_y & = & -{\dfrac{g}{2v_0^2\sin^2\theta}} & S_x^2 & - & \dfrac{1}{\tan\theta} & S_x & + & H\\
\downarrow & & \downarrow & \downarrow & & \downarrow & \downarrow & & \downarrow\\
y & = & a & x^2 & + & b & x & + & c
\end{array}
\]
we see that we obtained a function of the type Sy = f(Sx with the
coefficient a < 0, which indicates that the trajectory is a parabola with the concave downwards.
d) When the ball hits the ground its velocity has components in the x and y directions (Figure 3). The velocity in the x direction is given by expression (I), and the velocity in the y direction is obtained from expression (V), where the time is substituted by the value found in item (a)
\[
\begin{gather}
v_y=-v_0\cos\theta-\cancel{g}\left(\frac{-v_0\cos\theta+\sqrt{v_0^2\cos^2\theta+2gH}}{\cancel{g}}\right)\\[5pt]
v_y=-v_0\cos\theta+v_0\cos\theta-\sqrt{v_0^2\cos^2\theta+2gH}\\[5pt]
v_y=-\sqrt{v_0^2\cos^2\theta+2gH}
\end{gather}
\]
The speed of the ball will be given by the vector sum
\[
\begin{gather}
\vec v={\vec v}_x+{\vec v}_y
\end{gather}
\]
The magnitude is obtained by applying the Pythagorean Theorem
\[
\begin{gather}
v^2=v_x^2+v_y^2\\[5pt]
v^2=\left(v_0\sin\theta\right)^2+\left(-\sqrt{v_0^2\cos^2\theta+2gH}\right)^2\\[5pt]
v^2=v_0^2\sin^2\theta+v_0^2\cos^2\theta+2gH
\end{gather}
\]
factoring the term
\( v_0^2 \)
on the right-hand side of the equation
\[
\begin{gather}
v^2=v_0^2\;(\,\underbrace{{\sin}^2\theta+ \cos^2\theta}_1\,)+2gH
\end{gather}
\]
From the Trigonometry
\[
\begin{gather}
\sin^2\theta+\cos^2\theta=1
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v=\sqrt{v_0^2+2gH\;}}
\end{gather}
\]
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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 .