Question

Figure 1.60 shows the trajectory of a ball travelling through the air, affected by both gravity and air resistance.

Here are the positions of the ball at several successive times.

Location	t(s)	Position ( m)
A	0.0	(0,0,0)
B	1.0	(22.3,26.1,0)
C	2.0	(40.1,38.1,0)

a) What is the average velocity of the ball as it travels between location A and location B? b) If the ball continued to travel at the same average velocity during the next second, where would it be at the end of that second? (That is, where would it be at time t=2s )c) How does your prediction from part b) compare to the actual position of the ball at t=2s(location C)? If the predicted and the observed location of the ball are different, explain why?

Short answer

Answer

• The average velocity of the ball is$\left(22.3,26.1,0\right)m/s$
• b) the ball will be after$t=2s$atand$\left(44.6,52.2,0\right)m$
• c) the difference between the actual and the predicted result is$\left[-4.5,-14.1,0\right]m$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as

• The location of A at t=0 s is$\left(0,0,0\right)m$ .
• The location of B at t=1.0 is$\left(22.3,26.1,0\right)m$ .
• The location of C at t=2.0 is$\left(40.1,38.1,0\right)m$ .

Significance of Newton’s first law for the average velocity and distance 

This law states that an object will continue to move in a uniform motion unless it is resisted by an external object.

The equation of the average velocity and the equation of displacement give the average velocity and the distance of the ball.

Determination of the average velocity and distance

f the particle in the C point in the y-quadrant,

$y_B=$The displacement of the particle in the B point in the y -quadrant,

$t_{BC}=$The time difference of a particle from location B to location C

$v_{average.y}$=The average velocity of the particle in the Y quadrant

Substituting the values $t_{BC}=1s$, $v_{average.y}=26.1m/s$

a)From Newton’s first law, the formula of the average velocity of the ball is expressed as:

role="math" localid="1657711258477" $\underset{v_{average}}{\to }=\left(\frac{\left(x_A-x_B\right)}{∆t_{AB}},\frac{\left(y_A-y_B\right)}{∆t_{AB}},\frac{\left(z_A-z_B\right)}{∆t_{AB}}\right)$

Here, The time difference of a particle from location A to location B

$x_A,y_A,z_A=$The positions of the ball at A position in the x, y, and z coordinates

$x_B,y_B,z_B=$The positions of the ball at B position in the x, y, and z coordinates

Substituting the values $t_{AB}=1s-0s=1s$, role="math" localid="1657710564688" $x_A,y_A,z_A=\left(0,0,0\right)m$, and $x_B,y_B,z_B=\left(22.3,26.1,0\right)m$ in the above equation, we get

$$\begin{gathered} v_{average}=\left(\frac{\left(22.3m-0m\right)}{1s},\frac{\left(26.1m-0m\right)}{1s},\frac{\left(z0s-0s\right)}{1s}\right) \\ {v}_{average}=\left(22.3,26.1,0\right)m/s \end{gathered}$$

Thus, the average velocity of the ball is (22.3,26.1,0) m/s .

b) From Newton’s first law, the equation of displacement of the ball in the x-coordinate can be expressed as:

$x_C=x_B+v_{average,x}.∆t_{BC}$

Here, $x_C=$The displacement of the particle in the C point in the x-quadrant,

$x_B=$The displacement of the particle in the B point in the x-quadrant,

$t_{BC}=$The time difference of a particle from location B to location C

$v_{average,x}=$The average velocity of the particle in the X- quadrant

Substituting the values$t_{BC}=2s-1s=1s$ , $v_{average,x}=22.3m/s$and $x_B=22.3m$in the above equation, we get:

$$\begin{gathered} x_C=22.3m+\left(22.3m/s\right)\times 1s \\ {x}_C=44.6m \end{gathered}$$

Similarly, the equation of displacement of the ball in the y coordinate can be expressed as:

$y_C=y_B+v_{average,y}.∆t_{BC}$

Here, $y_C=$The displacement o$y_B=26.1m$ in the above equation, we get-

$$\begin{gathered} y_C=26.1m/s+\left(26.1m/s\right)\times 1s \\ {y}_C=52.2m \end{gathered}$$

As the value of the z coordinate is 0 , the displacement is also zero.
Thus, the ball will be after t = 2 s at (44.6,52.2,0)m .

c)

Hence, the difference between the predicted and the observed result can be expressed as:

${\left(x_C.y_C,z_C\right)}_{predicted}-{\left(x_C,y_C,z_C\right)}_{actual}$

Here,

${\left(x_C.y_C,z_C\right)}_{predicted}$=The predicted value of the ball and

${\left(x_C.y_C,z_C\right)}_{actual}=$The actual value of the ball

Substituting the values ${\left(x_C.y_C,z_C\right)}_{predicted}=\left(40.1,38.1,0\right)m$and${\left(x_C.y_C,z_C\right)}_{actual}=\left(44.6,52.2,0\right)m$ in the above equation, we get:

role="math" localid="1657715023925" $$\begin{gathered} \left[\left(40.1,-44.6\right),\left(38.1-52.2\right),\left(0-0\right)\right]m \\ =\left[-4.5,-14.1,0\right]m \end{gathered}$$

Thus, the difference between the actual and the predicted result is$=\left[-4.5,-14.1,0\right]m$ .

In the absence of the air resistance, the predicted and then the actual value may be the same as gravity is only acting in the y -direction and is also the only acting force. However, air resistance is also affecting our prediction of the value of $x_c$. The prediction of the y-coordinate is also affected by the gravitational force and also the atmospheric drag force.