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Vertical motion under gravity (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the accelerated motion under gravity. The questions are categorized in terms of the characterizing features of the subject matter :

  • Motion plots
  • Equal displacement
  • Equal time
  • Displacement in a particular second
  • Twice in a position
  • Collision in air

Motion plots

Example 1

Problem : A ball is dropped from a height of 80 m. If the ball looses half its speed after each strike with the horizontal floor, draw (i) speed – time and (ii) velocity – time plots for two strikes with the floor. Consider vertical downward direction as positive and g = 10 m / s 2 m / s 2 .

Solution : In order to draw the plot, we need to know the values of speed and velocity against time. However, the ball moves under gravity with a constant acceleration 10 m / s 2 m / s 2 . As such, the speed and velocity between strikes are uniformly increasing or decreasing at constant rate. It means that we need to know end values, when the ball strikes the floor or when it reaches the maximum height.

In the beginning when the ball is released, the initial speed and velocity both are equal to zero. Its velocity, at the time first strike, is obtained from the equation of motion as :

v 2 = 0 + 2 g h v = ( 2 g h ) = ( 2 x 10 x 80 ) = 40 m / s v 2 = 0 + 2 g h v = ( 2 g h ) = ( 2 x 10 x 80 ) = 40 m / s

Corresponding speed is :

| v | = 40 m / s | v | = 40 m / s

The time to reach the floor is :

t = v - u a = 40 - 0 10 = 4 s t = v - u a = 40 - 0 10 = 4 s

According to question, the ball moves up with half the speed. Hence, its speed after first strike is :

| v | = 20 m / s | v | = 20 m / s

The corresponding upward velocity after first strike is :

v = - 20 m / s v = - 20 m / s

Figure 1: The ball dropped from a height looses half its height on each strike with the horizontal surface.
Speed – time plot
 Speed – time plot  (vmg1a.gif)

Figure 2: The ball dropped from a height looses half its height on each strike with the horizontal surface.
Velocity – time plot
 Velocity – time plot  (vmg2.gif)

It reaches a maximum height, when its speed and velocity both are equal to zero. The time to reach maximum height is :

t = v - u a = 0 + 20 10 = 2 s t = v - u a = 0 + 20 10 = 2 s

After reaching the maximum height, the ball returns towards floor and hits it with the same speed with which it was projected up,

v = 20 m / s v = 20 m / s

Corresponding speed is :

| v | = 20 m / s | v | = 20 m / s

The time to reach the floor is :

t = 2 s t = 2 s

Again, the ball moves up with half the speed with which ball strikes the floor. Hence, its speed after second strike is :

| v | = 10 m / s | v | = 10 m / s

The corresponding upward velocity after first strike is :

v = - 10 m / s v = - 10 m / s

It reaches a maximum height, when its speed and velocity both are equal to zero. The time to reach maximum height is :

t = v - u a = 0 + 10 10 = 1 s t = v - u a = 0 + 10 10 = 1 s

Example 2

Problem : A ball is dropped vertically from a height “h” above the ground. It hits the ground and bounces up vertically to a height “h/3”. Neglecting subsequent motion and air resistance, plot its velocity “v” with the height qualitatively.

Solution : We can proceed to plot first by fixing the origin of coordinate system. Let this be the ground. Let us also assume that vertical upward direction is the positive y-direction of the coordinate system. The ball remains above ground during the motion. Hence, height (y) is always positive. Since we are required to plot velocity .vs. height (displacement), we can use the equation of motion that relates these two quantities :

v 2 = u 2 + 2 a y v 2 = u 2 + 2 a y

During the downward motion, u = 0, a = -g and displacement (y) is positive. Hence,

v 2 = - 2 g y v 2 = - 2 g y

This is a quadratic equation. As such the plot is a parabola as motion progresses i.e. as y decreases till it becomes equal to zero (see plot below y-axis).

Similarly, during the upward motion, the ball has certain velocity so that it reaches 1/3 rd of the height. It means that ball has a velocity less than that with which it strikes the ground. However, this velocity of rebound is in the upwards direction i.e positive direction of the coordinate system. In the nutshell, we should start drawing upward motion with a smaller positive velocity. The equation of motion, now, is :

v 2 = 2 g y v 2 = 2 g y

Again, the nature of the plot is a parabola the relation being a quadratic equation. The plot progresses till displacement becomes equal to y/3 (see plot above y-axis). The two plots should look like as given here :

Figure 3: The nature of plot is parabola.
Velocity – displacement plot
 Velocity – displacement plot  (vmg3.gif)

Equal displacement

Example 3

Problem : A particle is released from a height of “3h”. Find the ratios of time taken to fall through equal heights “h”.

Solution : Let t 1 t 1 , t 2 t 2 and t 3 t 3 be the time taken to fall through successive heights “h”. Then, the vertical linear distances traveled are :

h = 1 2 g t 1 2 2h = 1 2 g ( t 1 + t 2 ) 2 3h = 1 2 g ( t 1 + t 2 + t 3 ) 2 h = 1 2 g t 1 2 2h = 1 2 g ( t 1 + t 2 ) 2 3h = 1 2 g ( t 1 + t 2 + t 3 ) 2

t 1 2 : ( t 1 + t 2 ) 2 : ( t 1 + t 2 + t 3 ) 2 :: 1 : 2 : 3 t 1 2 : ( t 1 + t 2 ) 2 : ( t 1 + t 2 + t 3 ) 2 :: 1 : 2 : 3

t 1 : t 1 + t 2 : t 1 + t 2 + t 3 :: 1 : 2 : 3 t 1 : t 1 + t 2 : t 1 + t 2 + t 3 :: 1 : 2 : 3

In order to reduce this proportion into the one as required in the question, we carry out a general reduction of the similar type. In the end, we shall apply the result to the above proportion. Now, a general reduction of the proportion of this type is carried out as below. Let :

t 1 : t 1 + t 2 : t 1 + t 2 + t 3 :: a : b : c t 1 : t 1 + t 2 : t 1 + t 2 + t 3 :: a : b : c

t 1 a = t 1 + t 2 b = t 1 + t 2 + t 3 c t 1 a = t 1 + t 2 b = t 1 + t 2 + t 3 c

From the first two ratios,

t 1 a = t 1 + t 2 - t 1 b - a = t 2 b - a t 1 a = t 1 + t 2 - t 1 b - a = t 2 b - a

Similarly, from the last two ratios,

t 1 + t 2 b = t 1 + t 2 + t 3 - t 1 - t 2 c - b = t 3 c - b t 1 + t 2 b = t 1 + t 2 + t 3 - t 1 - t 2 c - b = t 3 c - b

Now, combining the reduced ratios, we have :

t 1 a = t 2 b - a = t 3 c - b t 1 a = t 2 b - a = t 3 c - b

t 1 : t 2 : t 3 :: a : b - a : c - b t 1 : t 2 : t 3 :: a : b - a : c - b

In the nutshell, we conclude that if

t 1 : t 1 + t 2 : t 1 + t 2 + t 3 :: a : b : c t 1 : t 1 + t 2 : t 1 + t 2 + t 3 :: a : b : c

Then,

t 1 : t 2 : t 3 :: a : b - a : c - b t 1 : t 2 : t 3 :: a : b - a : c - b

Applying this result as obtained to the question in hand,

t 1 : t 1 + t 2 : t 1 + t 2 + t 3 :: 1 : 2 : 3 t 1 : t 1 + t 2 : t 1 + t 2 + t 3 :: 1 : 2 : 3

We have :

t 1 : t 2 : t 3 :: 1 : 2 - 1 : 3 - 2 t 1 : t 2 : t 3 :: 1 : 2 - 1 : 3 - 2

Equal time

Example 4

Problem : Balls are successively dropped one after another at an equal interval from a tower. At the instant, 9 th 9 th ball is released, the first ball hits the ground. Which of the ball in series is at ¾ of the height of the tower?

Solution : Let “t” be the equal time interval. The first ball hits the ground when 9 th 9 th ball is dropped. It means that the first ball has fallen for a total time (9-1)t = 8t. Let n th n th ball is at 3/4th of the height of tower. Then,

For the first ball,

h = 1 2 g x 8 t 2 h = 1 2 g x 8 t 2

For the nth ball,

h - 3 h 4 = h 4 = 1 2 g x ( 9 - n ) 2 t 2 h - 3 h 4 = h 4 = 1 2 g x ( 9 - n ) 2 t 2

Combining two equations, we have :

h = 2 g x ( 9 - n ) 2 t 2 = 1 2 g x 8 t 2 = 32 g t 2 ( 9 - t ) 2 = 16 9 - n = 4 n = 5 h = 2 g x ( 9 - n ) 2 t 2 = 1 2 g x 8 t 2 = 32 g t 2 ( 9 - t ) 2 = 16 9 - n = 4 n = 5

Displacement in a particular second

Example 5

Problem : A ball is dropped vertically from a tower. If the vertical distance covered in the last second is equal to the distance covered in first 3 seconds, then find the height of the tower. Consider g = 10 m / s 2 m / s 2 .

Solution : Let us consider that the ball covers the height y n y n in n th n th second. Then, the distance covered in the n th n th second is given as :

y n = u + g 2 x ( 2 n - 1 ) = 0 + 10 2 x ( 2 n - 1 ) = 10 n - 5 y n = u + g 2 x ( 2 n - 1 ) = 0 + 10 2 x ( 2 n - 1 ) = 10 n - 5

On the other hand, the distance covered in first 3 seconds is :

y 3 = u t + 1 2 g t 2 = 0 + 1 2 x 10 x 3 2 = 45 m y 3 = u t + 1 2 g t 2 = 0 + 1 2 x 10 x 3 2 = 45 m

According to question,

y n = y 3 y n = y 3

10 n - 5 = 45 n = 5 s 10 n - 5 = 45 n = 5 s

Therefre, the height of the tower is :

y = 1 2 x 10 x 5 2 = 125 m y = 1 2 x 10 x 5 2 = 125 m

Twice in a position

Example 6

Problem : A ball, thrown vertically upward from the ground, crosses a point “A” in time “ t 1 t 1 ”. If the ball continues to move up and then return to the ground in additional time “ t 2 t 2 ”, then find the height of “A” from the ground. .

Solution : Let the upward direction be the positive reference direction. The displacement equation for vertical projection from the ground to the point “A”, is :

h = u t 1 + 1 2 g t 1 2 h = u t 1 + 1 2 g t 1 2

In this equation, the only variable that we do not know is initial velocity. In order to determine initial velocity, we consider the complete upward motion till the ball reaches the maximum height. We know that the ball takes half of the total time to reach maximum height. It means that time for upward motion till maximum height is :

t 1 2 = t 1 + t 2 2 t 1 2 = t 1 + t 2 2

As we know the time of flight, final velocity and acceleration, we can know initial velocity :

0 = u - g t 1 2 u = g t 1 2 = ( t 1 + t 2 ) g 2 0 = u - g t 1 2 u = g t 1 2 = ( t 1 + t 2 ) g 2

Substituting this value, the height of point “A” is :

h = u t 1 + 1 2 g t 1 2 h = ( t 1 + t 2 ) g 2 x t 1 - 1 2 g t 1 2 h = g t 1 2 + g t 1 t 2 - g t 1 2 2 h = g t 1 t 2 2 h = u t 1 + 1 2 g t 1 2 h = ( t 1 + t 2 ) g 2 x t 1 - 1 2 g t 1 2 h = g t 1 2 + g t 1 t 2 - g t 1 2 2 h = g t 1 t 2 2

Example 7

Problem : A ball, thrown vertically upward from the ground, crosses a point “A” at a height 80 meters from the ground. If the ball returns to the same position after 6 second, then find the velocity of projection. (Consider g = 10 m / s 2 m / s 2 ).

Solution : Since two time instants for the same position is given, it is indicative that we may use the displacement equation as it may turn out to be quadratic equation in time “t”. The displacement, “y”, is (considering upward direction as positive y-direction) :

y = u t + 1 2 g t 2 80 = u t + 1 2 x - 10 t 2 = u t - 5 t 2 5 t 2 - u t + 80 = 0 y = u t + 1 2 g t 2 80 = u t + 1 2 x - 10 t 2 = u t - 5 t 2 5 t 2 - u t + 80 = 0

We can express time “t”, using quadratic formulae,

t 2 = - ( - u ) + { ( - u ) 2 - 4 x 5 x 80 } 2 x 5 = u + ( u 2 - 1600 ) 10 t 2 = - ( - u ) + { ( - u ) 2 - 4 x 5 x 80 } 2 x 5 = u + ( u 2 - 1600 ) 10

Similarly,

t 1 = u - ( u 2 - 1600 ) 10 t 1 = u - ( u 2 - 1600 ) 10

According to question,

t 2 - t 1 = ( u 2 - 1600 ) 5 6 = ( u 2 - 1600 ) 5 ( u 2 - 1600 ) = 30 t 2 - t 1 = ( u 2 - 1600 ) 5 6 = ( u 2 - 1600 ) 5 ( u 2 - 1600 ) = 30

Squaring both sides we have,

u 2 - 1600 = 900 u = 50 m / s u 2 - 1600 = 900 u = 50 m / s

Collision in air

Example 8

Problem : A ball “A” is dropped from a height, “h”, when another ball, “B” is thrown upward in the same vertical line from the ground. At the instant the balls collide in the air, the speed of “A” is twice that of “B”. Find the height at which the balls collide.

Solution : Let the velocity of projection of “B” be “u” and let the collision occurs at a fraction “a” of the height “h” from the ground. We should pause and make mental note of this new approach to express an intermediate height in terms of fraction of total height (take the help of figure).

Figure 4: The balls collide as they move in same vertical line.
Collision in the mid air
 Collision in the mid air  (vmg4.gif)

Adhering to the conventional approach, we could have denoted the height at which collision takes place. For example, we could have considered collision at a vertical displacement y from the ground. Then displacements of two balls would have been :

“y” and “h – y”

Obviously, the second expression of displacement is polynomial of two terms because of minus sign involved. On the other hand, the displacements of two balls in terms of fraction, are :

“ha” and “(1-a)h”

The advantage of the second approach (using fraction) is that displacements are stated in terms of the product of two quantities. The variable “h” appearing in each of the terms cancel out if they appear on ether side of the equation and we ultimately get equation in one variable i.e. “a”. In the nutshell, the approach using fraction reduces variables in the resulting equation. This point will be highlighted at the appropriate point in the solution to appreciate why we should use fraction?

Now proceeding with the question, the vertical displacement of “B” is ah and that of “A” is (1-a)h. We observe here that each of the balls takes the same time to cover respective displacements. Using equation of constant acceleration in one dimension (considering downward vertical direction positive),

For ball “A”,

( 1 - a ) h = 1 2 g t 2 ( 1 - a ) h = 1 2 g t 2

For ball “B”,

- a h = - u t + 1 2 g t 2 - a h = - u t + 1 2 g t 2

The value of “t” obtained from the equation of ball “A” is :

t = { 2 ( 1 - a ) h g } t = { 2 ( 1 - a ) h g }

Substituting in the equation of ball “B”, we have :

- a h = - u { 2 ( 1 - a ) h g } + ( 1 - a ) h u { 2 ( 1 - a ) h g } = h u = { g h 2 ( 1 - a ) } - a h = - u { 2 ( 1 - a ) h g } + ( 1 - a ) h u { 2 ( 1 - a ) h g } = h u = { g h 2 ( 1 - a ) }

Now, according to the condition given in the question,

v A = 2 v B v A 2 = 4 v B 2 v A = 2 v B v A 2 = 4 v B 2

Using equations of motion for constant acceleration, we have :

2 g ( 1 - a ) h = 4 ( u 2 - 2 g a h ) 2 g ( 1 - a ) h = 4 { g h 2 ( 1 - a ) - 2 g a h } ( 1 - a ) = { 1 ( 1 - a ) - 4 a } 2 g ( 1 - a ) h = 4 ( u 2 - 2 g a h ) 2 g ( 1 - a ) h = 4 { g h 2 ( 1 - a ) - 2 g a h } ( 1 - a ) = { 1 ( 1 - a ) - 4 a }

Recall our discussion on the use of fraction. We can write down the corresponding steps for conventional method to express displacements side by side and find that the conventional approach will lead us to an equation as :

( h - y ) = { h ( h - y ) - 4 y } ( h - y ) = { h ( h - y ) - 4 y }

Evidently, this is an equation in two variables and hence can not be solved. It is this precise limitation of conventional approach that we switched to fraction approach to get an equation in one variable,

( 1 - a ) = { 1 ( 1 - a ) - 4 a } ( 1 - a ) = { 1 ( 1 - a ) - 4 a }

( 1 - a ) 2 = 1 - 4 a ( 1 - a ) 1 + a 2 - 2 a = 1 - 4 a + a 2 3 a 2 - 2 a = 0 a ( 3 a - 2 ) = 0 a = 0 , 2 3 ( 1 - a ) 2 = 1 - 4 a ( 1 - a ) 1 + a 2 - 2 a = 1 - 4 a + a 2 3 a 2 - 2 a = 0 a ( 3 a - 2 ) = 0 a = 0 , 2 3

Neglecting, a =0,

a = 2 3 a = 2 3

The height from the ground at which collision takes place is :

y = a h = 2 h 3 y = a h = 2 h 3

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