Free Falling Bodies

A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward or downward and those released from rest are all falling freely once they are released. Any freely falling object experiences acceleration directed downward, regardless of its initial motion.

We shall denote the magnitude of the free-fall acceleration by the symbol g. The value of g near the Earth’s surface decreases with increasing altitude. Furthermore, slight variations in g occur with changes in latitude. It is common to define “up” as the +y direction and to use y as the position variable in the kinematic equations. At the Earth’s surface, the value of g is approximately 9.80 m/s2.

If we neglect air resistance and assume that the free-fall acceleration does not vary with altitude over short vertical distances, then the motion of a freely falling object moving vertically is equivalent to motion in one dimension under constant acceleration. We always choose ay=-g =9.80m/s2, where the negative sign means that the acceleration of a freely falling object is downwards. Therefore, the equations of motion developed for objects moving with constant acceleration can be applied.

Down ward motion

i.

ii.

iii.

Upward motion

i.

ii.

iii.

Examples

1) A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down, as shown in Figure below. Using tA=0 as the time the stone leaves the thrower’s hand at position A, determine the;

(a) Time at which the stone reaches its maximum height,

(b) Maximum height,

(c) Time at which the stone returns to the height from which it was thrown,

(d) Velocity of the stone at this instant, and

Solution a) Initial velocity=20 m/s

Final velocity=0 To calculate the time t B at which the stone reaches maximum height, we use equation

b)

c) When the stone is back at the height from which it was thrown the y coordinate is again zero.

This is a quadratic equation and so has two solutions One solution is t=0

corresponding to the time the stone starts its motion. The other solution is t=4.08s which is the solution we are after.

d) Velocity at t=4.08s

but displacement at point of throw is zero. Hence

+20m/s is the initial velocity of the body moving upwards while -20m/s is the velocity of the falling object at the point of throw.The body hits the point of throw with a velocity equal in magnitude but opposite in direction to the initial velocity.

2) A ball thrown vertically upward is caught by the thrower after 20.0 s. Find (a) the initial velocity of the ball and,(b) The maximum height it reaches.

3) A ball is thrown vertically upward from the ground with an initial speed of 15.0 m/s. (a) How long does it take the ball to reach its maximum height? (b) What is its maximum height? (c) Determine the velocity and acceleration of the ball at t = 2.00 s.

4) A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upward from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.

Motion in two dimensionsThe position, velocity and acceleration vectorsLet describe the position of a particle by its position vector r, drawn from the origin of some coordinate system to the particle located in the xy plane, as in Figure below.

At time to the particle is at point A, and at some later time tf it is at point B. The path from A to B is not necessarily a straight line. As the particle moves from A to B in the time interval its position vector changes from ro to rf. displacement is a vector quantity and the displacement of the particle is the difference between its final position and its initial position. We now formally define the displacement vector for the particle as being the difference between its final position vector rf and its initial position vector ro:

We define the velocity of a particle during the time interval as the displacement of the particle divided by that time interval:

Multiplying or dividing a vector quantity by a scalar quantity changes only the magnitude of the vector, not its direction. Because displacement is a vector quantity and the time interval is a scalar quantity, we conclude that the average velocity is a vector quantity directed along .

The instantaneous velocity v is defined as the limit of the average velocity as approaches

zero:

That is, the instantaneous velocity equals the derivative of the position vector with respect to time. The direction of the instantaneous velocity vector at any point in a particle’s path is along a line tangent to the path at that point and in the direction of motion. The magnitude of the instantaneous velocity vector is called the speed.

As a particle moves from one point to another along some path, its instantaneous velocity vector changes from vo at time to to v at time tf . Knowing the velocity at these points allows us to determine the average acceleration of the particle.The average acceleration of a particle as it moves from one position to another is defined as the change in the instantaneous velocity vector divided by the time during which that change occurred:

When the average acceleration of a particle changes during different time intervals, it is useful to define its instantaneous acceleration a. The instantaneous acceleration a is defined as the limiting value of the ratio

as approaches zero:

NB: It is important to recognize that various changes can occur when a particle accelerates. i. The magnitude of the velocity vector (the speed) may change with time as in straight-line

(one-dimensional) motion. ii. The direction of the velocity vector may change with time even if its magnitude (speed)

remains constant, as in curved-path (two-dimensional) motion. iii. Both the magnitude and the direction of the velocity vector may change simultaneously.

Two-Dimensional Motion with Constant AccelerationLet us consider two-dimensional motion during which the acceleration remains constant in both magnitude and direction. The position vector for a particle moving in the xy plane can be written

……………………………….1where x, y, and r change with time as the particle moves while i and j remain constant. If the position vector is known, the velocity of the particle can be obtained by getting a differential of equation 1 above.

…………………………………2

Because is assumed constant, its components ax and ay also are constants. Therefore, we can apply the equations of kinematics to the x and y components of the velocity vector. Substituting

and into Equation 2 to determine the final velocity at any time t, we obtain

Quiz 4.1

This result states that the velocity of a particle at some time t equals the vector sum of its initial velocity v0 and the additional velocity acquired in the time t as a result of constant acceleration.

Similarly, we know that the x and y coordinates of a particle moving with constant acceleration are

and respectively.

Substituting these expressions into Equation 1(and labeling the final position vector rf) gives

This equation tells us that the position vector rf is the vector sum of original position r0, a

displacement v0t arising from the initial velocity of the particle and a displacement resulting

from the uniform acceleration of the particle.

Projectile Motion

Projectile is a body thrown with an initial velocity in the vertical plane and then it moves in two dimensions under the action of gravity. Its motion is called projectile motion. The path of a projectile is called its trajectory.

Examples: a golf ball in flight, a bullet fired from a rifle and a jet of water from a hole near the bottom of a water tank.

Projectile motion is a case of two-dimensional motion. In dealing with projectile motion two assumptions are made

The free fall acceleration g is constant over the range of motion and is directed down ward

The effect of air resistance is negligible.

With these assumptions, we find that the path of a projectile is a parabola. To show that the trajectory of a projectile is a parabola, let us choose our reference frame such that the y direction is vertical and positive is upward. Because air resistance is neglected, we know that and that Furthermore, let us assume that at t = 0, the projectile leaves the origin with initial velocity , as shown in Figure 1. The vector makes an angle with the horizontal, where is the angle at which the projectile leaves the origin.

From the definitions of the cosine and sine functions we have

and therefore, the table below gives a summary of y and x components of

this motion.

X axis Y axis

1. Component of initial velocity along x-axis. 1. Component of initial velocity along y-axis.

2. Component of velocity along the x-axis at any instant t.

This means that the horizontal component of velocity does not change throughout the projectile motion.

2. Component of velocity along the y-axis at any instant t.

3. The displacement along x-axis at any instant t

»

3. The displacement along y-axis at any instant t

»

 Equation of Trajectory (Path of projectile)

At any instant t

»

Also,

Substituting for t

» …………………………….1

This equation is of the form where 'a' and 'b are constants. This is the equation of a

parabola. This equation of motion is valid for launch angles in the range .

Net velocity of the body at any instant of time t

……………………………………2

Where is the angle that the resultant velocity (v) makes with the horizontal at

any instant?

Time to reach the maximum height

Angular Projectile motion is symmetrical about the highest point. The object will reach the highest point in time .At the highest point, the vertical component of velocity becomes equal to zero.

At

Time of Flight T

Angular Projectile motion is symmetrical about the highest point. The object will reach the

highest point in time .At the highest point, the vertical component of velocity becomes

equal to zero.

At

» ……………………………………..3

Maximum height H

Let consider figure 2 below

Equation for vertical distance (y component)

At

»

substituting the value of T from equation 3 we get

=

»

Range R

Range is the total horizontal distance covered during the time of flight. From equation for horizontal motion, when thus;

Using the trigonometric identity 2sinØcosØ= sin2Ø

»

The maximum value of R from Equation above is . This result follows from the fact that

the maximum value of is 1, which occurs when . Therefore, R is a maximum when . In addition, a point having Cartesian coordinates (R, 0) can be reached by using either one of two complementary values of , such as 75° and 15° as shown in the figure below. Of course, the maximum height and time of flight for one of these values of are different from the maximum height and time of flight for the complementary value.

Examples