Concept explainers
A projectile is launched from the point (x = 0, y = 0), with velocity
(a)
The values of the projectile distance
Answer to Problem 72AP
The table for the values of
|
|
0 | 0 |
1 | 45.7 |
2 | 82.0 |
3 | 109 |
4 | 127 |
5 | 136 |
6 | 138 |
7 | 133 |
8 | 124 |
9 | 117 |
10 | 120 |
Explanation of Solution
Given info: The initial position of the projectile is
The value of the acceleration due to gravity is
The formula to calculate the
Here,
In a vertical projectile the acceleration in
Substitute
Thus, the
The formula to calculate the
Here,
For the vertical projectile the acceleration in
Substitute
Thus, the
The formula to calculate the magnitude of the position vector is,
Substitute
For
|
|
0 | 0 |
1 | 45.7 |
2 | 82.0 |
3 | 109 |
4 | 127 |
5 | 136 |
6 | 138 |
7 | 133 |
8 | 124 |
9 | 117 |
10 | 120 |
Conclusion:
Therefore, the table for the values of
|
|
0 | 0 |
1 | 45.7 |
2 | 82.0 |
3 | 109 |
4 | 127 |
5 | 136 |
6 | 138 |
7 | 133 |
8 | 124 |
9 | 117 |
10 | 120 |
(b)
To show: The distance is maximum when the position vector is perpendicular to the velocity.
Answer to Problem 72AP
The distance is maximum when the position vector is perpendicular to the velocity.
Explanation of Solution
Given info: The initial position of the projectile is
The velocity vector tells about the change in the position vector. If the velocity vector at particular point has a component along the position vector and the velocity vector makes an angle less than
For the position vector to be maximum the distance from the origin must be momentarily at rest or constant and the only possible situation for this is that the velocity vector makes an angle
Conclusion:
Therefore, the distance is maximum when the position vector is perpendicular to the velocity vector.
(c)
The magnitude of the maximum displacement.
Answer to Problem 72AP
The magnitude of the maximum displacement is
Explanation of Solution
Given info: The initial position of the projectile is
The expression for the position vector is,
Square both the sides of the above equation.
Differentiate the above expression with respect to
For the maxima condition to calculate the value the value of
Further solve the above expression for
Further solving the above quadratic equation the values of
From the table in Part (a) it is evident that after
Thus, the maximum value of
Substitute
Conclusion:
Therefore, the maximum distance is
(d)
To write: The explanation for the method used in part (c) calculation.
Answer to Problem 72AP
The maxima and minima condition is used to calculate the maximum magnitude of the position vector.
Explanation of Solution
Given info: The initial position of the projectile is
The maximum or minimum value of any function is easily calculated using the Maxima and Minima condition.
For the part (c) first calculate the critical points by equating the differential of
The value of
Substitute the maximum value of
Conclusion:
Therefore, the maxima and minima condition is used to calculate the maximum magnitude of the position vector.
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Chapter 4 Solutions
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
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