Chapter 3, Problem 31QAP

To determine

(a)

To Draw:

A graph of v_x versus time for a projectile and identify the points where the object reaches its highest point and where it hits the ground at the end of its flight.

Explanation of Solution

Introduction:

Consider an object that undergoes a projectile motion. The object is projected with a velocity v at an angle θ to the horizontal. The velocity has an initial horizontal component ${\left(v_x\right)}_0$ and a vertical component ${\left(v_y\right)}_0$.

For the purpose of drawing a graph showing the variation of v_x with time, assume the value of ${\left(v_x\right)}_0$ to have a value 25 m/s. The horizontal component of the projectile's velocity remains constant since no force acts on the projectile in the horizontal direction.

On a spread sheet insert the values as shown and draw a graph as shown.

t in s	vx in m/s
0	25
1	25
2	25
3	25
4	25
5	25
6	25
7	25

Conclusion:

The graph is a straight line parallel to the time axis. From the above graph, it is not possible to identify the points where the object reaches its maximum height or when it reaches the ground, since the graph shows no variation of the horizontal component with time.

To determine

(b)

To Draw:

A graph of v_y versus time for a projectile and identify the points where the object reaches its highest point and where it hits the ground at the end of its flight.

Explanation of Solution

Introduction:

Consider an object that undergoes a projectile motion. The object is projected with a velocity v at an angle θ to the horizontal. The velocity has an initial horizontal component ${\left(v_x\right)}_0$ and a vertical component ${\left(v_y\right)}_0$. The vertical component of the velocity is directed upwards.

The projectile after it is launched is under the action of the acceleration of free fall. The vertical component of the projectile varies under the action of the acceleration due to gravity g.

The expression which shows the variation of v_y with time is given by,

  $v_y={\left(v_y\right)}_0+gt$

The acceleration due to gravity g is directed downwards, opposite to the direction of the vertical component of the velocity at the moment of launch. Therefore, it is given a value $\left(-9.81{\text{ m/s}}^2\right)$.

Assume, the value of ${\left(v_y\right)}_0$ to be 30 m/s.

On a spreadsheet, use the following equation to plot the variation of v_y with time.

  $v_y=\left(30\text{ m/s}\right)+\left(-9.81{\text{ m/s}}^2\right)t$

t in s	vy in m/s
0	30
0.1	29.019
0.2	28.038
0.3	27.057
0.4	26.076
0.5	25.095
0.6	24.114
0.7	23.133
0.8	22.152
0.9	21.171
1	20.19
1.1	19.209
1.2	18.228
1.3	17.247
1.4	16.266
1.5	15.285
1.6	14.304
1.7	13.323
1.8	12.342
1.9	11.361
2	10.38
2.1	9.399
2.2	8.418
2.3	7.437
2.4	6.456
2.5	5.475
2.6	4.494
2.7	3.513
2.8	2.532
2.9	1.551
3	0.57
3.1	-0.411
3.2	-1.392
3.3	-2.373
3.4	-3.354
3.5	-4.335
3.6	-5.316
3.7	-6.297
3.8	-7.278
3.9	-8.259
4	-9.24
4.1	-10.221
4.2	-11.202
4.3	-12.183
4.4	-13.164
4.5	-14.145
4.6	-15.126
4.7	-16.107
4.8	-17.088
4.9	-18.069
5	-19.05
5.1	-20.031
5.2	-21.012
5.3	-21.993
5.4	-22.974
5.5	-23.955
5.6	-24.936
5.7	-25.917
5.8	-26.898
5.9	-27.879
6	-28.86
6.1	-29.841
6.2	-30.822

From the graph, the following points are noted:

1.The maximum positive value of v_y is at point O, which the point at which it is launched.

2. As the object rises, the vertical component of its velocity reduces under the action of acceleration due to gravity g. When it reaches the maximum height, this component attains a value zero. This is seen from the graph as point A. The time taken by this projectile to reach its maximum height is seen to be equal to 3.05 s.

3. As the object descends from the position of maximum height, the vertical component of its velocity is directed downwards and hence it has a negative value. Its value increases as it moves downwards, since now, the vertical component v_y and the acceleration due to gravity are both directed downwards. When it hits the ground, the vertical component v_y has a magnitude of 30 m/s and it is directed downwards, hence negative. Point B represents the point when it reaches the ground. The total time of flight from the graph is 6.1 s.

Conclusion:

The object with a vertical component of magnitude 30 m/s, reaches the position of maximum height in time 3.05 s and reaches the ground after 6.1 s. The graph varies linearly with time. The gradient of the graph is equal to g which has a value $\left(-9.81{\text{ m/s}}^2\right)$.

To determine

(c)

To Draw:

A graph showing the variation of a_x with time.

Explanation of Solution

Introduction:

A projectile travels a parabolic path under the action of acceleration due to gravity, if no air resistance acts on it.

The only force acting on the projectile is the gravitational force, which is directed downwards, towards the center of the Earth. This force is constant at points close to the surface of the earth. Since it is directed towards the ground, it has no component along the horizontal direction. Hence, $a_x=0$ at all values of time.

On a spreadsheet, use $a_x=0$ and plot a graph as shown below.

tin s	ax in m/s2
0	0
1	0
2	0
3	0
4	0
5	0
6	0
7	0

Conclusion:

Since the only force acting on the projectile is the gravitational force and it has no component along the horizontal direction, the horizontal component of its acceleration is zero at all times, as shown in the graph.

To determine

(d)

To Draw:

A graph showing the variation of a_y with time.

Explanation of Solution

Introduction:

A projectile travel in a parabolic path under the action of acceleration due to gravity. If no air resistance acts on the object, the only force acting on it is the gravitational force. Hence the vertical component of the object's acceleration is $a_y=g$, which is equal to the acceleration due to gravity.

The gravitational force is a constant at points close to the surface of the earth. The acceleration due to gravity has a constant value of $\left(-9.81{\text{ m/s}}^2\right)$. The negative sign shows that this acceleration is directed downwards.

On a spreadsheet, use the expression $a_y=-9.81{\text{ m/s}}^2$ and plot a graph between a_y and t.

tin s	ay in m/s2
0	-9.81
1	-9.81
2	-9.81
3	-9.81
4	-9.81
5	-9.81
6	-9.81
7	-9.81

Conclusion:

The graph of a_y vs time is a straight line parallel to the time axis and has a constant value of $\left(-9.81{\text{ m/s}}^2\right)$.