Vertical motion under gravity *

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1 OpenStax-CNX module: m Vertical motion under gravity * Sunil Kumar Singh This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 2.0 Vertical motion under gravity is a specic case of one dimensional motion with constant acceleration. Here, acceleration is always directed in vertically downward direction and its magnitude is "g". As the force due to gravity may be opposite to the direction of motion, there exists the possibility that the body under force of gravity reverses its direction. It is, therefore, important to understand that the quantities involved in the equations of motion may evaluate to positive or negative values with the exception of time (t). We must appropriately assign sign to various inputs that goes into the equation and correctly interpret the result with reference to the assumed positive direction. Further, some of them evaluate to two values one for one direction and another of reversed direction. As pointed out earlier in the course, we must also realize that a change in reference direction may actually change the sign of the attributes, but their physical interpretation remains same. What it means that an attribute such as velocity, for example, can be either 5 m/s or -5 m/s, conveying the same velocity. The interpretation must be done with respect to the assigned positive reference direction. 1 Velocity Let us analyze the equation "v = u + at" for the vertical motion under gravity with the help of an example. We consider a ball thrown upwards from ground with an initial speed of 30 m/s. In the frame of reference with upward direction as positive, u = 30 m / s and a = g = 10 m / s 2 * Version 1.12: Oct 3, :12 am

2 OpenStax-CNX module: m Vertical motion under gravity Figure 1: The ball reaches maximum height when its velocity becomes zero Putting this value in the equation, we have : v = t The important aspect of this equation is that velocity evaluates to both positive and negative values; positive for upward motion and negative for downward motion. The nal velocity (v) is positive for t < 3 seconds, zero for t = 3 seconds and negative for t > 3 seconds. The total time taken for the complete up and down journey is 3 (for upward motion) + 3 (for downward motion) = 6 seconds. The velocities of the ball at successive seconds are : Time (t) Final velocity (v) in seconds in m/s The corresponding velocity time plot looks like as shown in the gure.

3 OpenStax-CNX module: m Velocity time plot Figure 2 We notice following important characteristics of the motion : 1: The velocity at the maximum height is zero (v=0). 2: The time taken by the ball to reach maximum height is obtained as : For v = 0, v = u + at = u gt = 0 u = gt t = u g 3: The ball completely regains its speed when it returns to ground, but the motion is directed in the opposite direction i.e. v = 4: The time taken for the complete round trip is : u For v = u, v = u + at = u gt u = u gt t = 2u g

4 OpenStax-CNX module: m The time taken for the complete journey is twice the time taken to reach the maximum height. It means that the ball takes equal time in upward and downward journey. Thus, the total motion can be considered to be divided in two parts of equal duration. 5: The velocity of the ball is positive in the rst half of motion; Zero at the maximum height; negative in the second of the motion. 6: The velocity is decreasing all through the motion from a positive value to less positive value in the rst half and from a less negative value to more negative value in the second half of the motion. This renders acceleration to be always negative (directed in -y direction), which is actually the case. 7: The velocity (positive) and acceleration (negative) in the rst part are opposite in direction and the resulting speed is decreasing. On the other hand, the velocity (negative) and acceleration (negative) in the second part are in the same direction and the resulting speed is increasing. 2 Displacement and distance Let us analyze the equation x = x 2 x 1 = ut at2 for the vertical motion under gravity with the help of earlier example. If we choose initial position as the origin, then x 1 = 0, x = x 2 = x ( say ) and x = ut at2, where x denotes position and displacement as well. In the frame of reference with upward direction as positive, Putting these values in the equation, we have : u = 30 m / s and a = g = 10 m / s 2 x = 30t 5t 2 The important aspect of this equation is that it is a quadratic equation in time t. This equation yields two values of time t for every position and displacement. This outcome is in complete agreement with the actual motion as the ball reaches a given position twice (during upward and downward motion). Only exception is point at the maximum height, which is reached only once. We have seen earlier that ball reaches maximum height at t = 3 s. Therefore, maximum height,h, is given as : H = 30 X 3 5 X 9 = 45 m The displacement values for the motion at successive seconds are : Time (t) in seconds ut 5txt Displacement in meters or position (x) The corresponding displacement time plot looks like as shown in the gure.

5 OpenStax-CNX module: m Displacement time plot Figure 3 We notice following important characteristics of the motion : 1: The ball retraces every position during motion except the point at maximum height. 2: The net displacement when ball return to initial position is zero. Thus, the total time of journey (T) is obtained using displacement, x = 0, For x = 0, x = ut at2 = ut 1 2 gt2 = 0 2uT gt 2 = 0 T = 2u g Here, we neglect T = 0, which corresponds to initial position. 3: The x in equation x = ut at2 denotes displacement and not distance. Hence, it is not possible to use this equation directly to obtain distance, when motion is not unidirectional. Let us answer the question with respect to the motion of the ball under consideration : what is the distance traveled in rst 4 seconds? Obviously, the ball travels 30 m in the upward direction to reach maximum height in 3 seconds and then it travels 5 m in the 4 th second in downward direction. Hence, the total distance traveled is = 50 m in 4 s. This means that we need to apply the equation of motion in two parts : one for the upward motion and the second for the downward motion. Thus, we nd displacement for each segment of the motion and then we can add their magnitude to obtain distance. The distance values for the motion at successive seconds are :

6 OpenStax-CNX module: m Time (t) ut 5t*t Displacement Distance in seconds or position in meters (x) in meters Distance time plot Figure 4 Example 1: Constant acceleration Problem : A balloon starts rising from the ground with an acceleration of 1.25 m/s. After 8 second, a stone is released from the balloon. Starting from the release of stone, nd the displacement and distance traveled by the stone on reaching the ground. Also, nd the time taken to reach the ground (take g = 10 m / s 2 ). Solution : This question raises few important issues. First the rise of balloon is at a constant acceleration of 1.25 m / s 2. This acceleration is the measured acceleration, which is net of the downward acceleration due to gravity. This means that the balloon rises with this net vertical acceleration of 1.25 m / s 2 in the upward direction.

7 OpenStax-CNX module: m Motion under gravity Figure 5 Here, u = 0; a = 1.25 m / s 2 ; t = 8 s. Let the balloon rises to a height h during this time, then (considering origin on ground and upward direction as positive) the displacement of the balloon after 8 seconds is : y = ut at2 = 0 x x 1.25 x 8 2 = 40 m Now, we know that, the body released from moving body acquires the velocity but not the acceleration of the container body under motion. The velocity of the balloon at the instant of separation is equal to the velocity of the balloon at that instant. v = u + at = x 8 = 10 m / s Thus, this is the initial velocity of the stone and is directed upward as that of the velocity of balloon. Once released, the stone is acted upon by the force of gravity alone. The role of the acceleration of the balloon is over. Now, the acceleration for the motion of stone is equal to the acceleration due to gravity, g. The path of motion of the stone is depicted in the gure. Stone rises due to its initial upward velocity to a certain height above 40 m where it was released till its velocity is zero. From this highest vertical point, the stone falls freely under gravity and hits the ground.

8 OpenStax-CNX module: m : In order to describe motion of the stone once it is released, we realize that it would be easier for us if we shift the origin to the point where stone is released. Considering origin at the point of release and upward direction as positive as shown in the gure, the displacement during the motion of stone is : y = OB = - 40 m Motion under gravity Figure 6 2: Distance, on the other hand, is equal to : s = OA + AO + OB = 2 OA + OB In order to obtain, OA, we consider this part of rectilinear motion (origin at the point of release and upward direction as positive as shown in the gure). Here, u = 10 m/s; a = -10 m / s 2 and v = 0. Applying equation of motion, we have : v 2 = u 2 + 2ay y = v2 u 2 2a = x 10 = 5 m Hence, OA = 5 m, and distance is : s = 2 OA + OB = 2 x = 50 m 3: The time of the journey of stone after its release from the balloon is obtained using equation of motion (origin at the point of release and upward direction as positive as shown in the gure). Here, u = 12 m/s; a = -10 m / s 2 and y = - 40 m.

9 OpenStax-CNX module: m y = ut at2 40 = 10t 0.5 x 10t 2 5t 2 10t 40 = 0 This is a quadratic equation in t. Its solution is : 5t t 20t 40 = 0 5t ( t + 2 ) 20 ( t + 2 ) = 0 ( 5t 20 ) ( t + 2 ) = 0 4, 2 s As negative value of time is not acceptable, time to reach the ground is 4s. note: It is important to realize that we are at liberty to switch origin or direction of reference after making suitable change in the sign of attributes. 3 Position We use the equation x = x 2 x 1 = ut at2 normally in the context of displacement, even though the equation is also designed to determine initial ( x 1 ) or nal position ( x 2 ). In certain situations, however, using this equation to determine position rather than displacement provides more elegant adaptability to the situation. Let us consider a typical problem highlighting this aspect of the equation of motion. Example 2 Problem : A ball is thrown vertically from the ground at a velocity 30 m/s, when another ball is dropped along the same line, simultaneously from the top of tower 120 m in height. Find the time (i) when the two balls meet and (ii) where do they meet. Solution : This question puts the position as the central concept. In addition to equal time of travel for each of the balls, rhe coordinate positions of the two balls are also same at the time they meet. Let this position be y. Considering upward direction as the positive reference direction, we have :

10 OpenStax-CNX module: m Vertical motion under gravity Figure 7: The balls have same coordinate value when they meet. For ball thrown from the ground : u = 30 m / s, a = 10 m / s 2, y 1 = 0, y 1 = y y 2 y 1 = ut at2 y 0 = 30t x t2 y = 30t 5 x t 2 (7) For ball dropped from the top of the tower : u = 0 m / s, a = 10 m / s 2, y 1 = 120, y 2 = y y 2 y 1 = ut at2 y 120 = 5xt 2 (7) Now, deducting equation (2) from (1), we have :

11 OpenStax-CNX module: m t = 120 t = Putting this value in equation 1, we have : y = 30x4 5x4 2 4 s = = 40 m One interesting aspect of this simultaneous motion of two balls is that the ball dropped from the tower meets the ball thrown from the ground, when the ball thrown from the ground is actually returning from after attaining the maximum height in 3 seconds. For maximum height of the ball thrown from the ground, Vertical motion under gravity Figure 8: When returning from the maximum height, the ball thrown up from the ground is hit by the ball dropped from towers. u = 30 m / s, a = 10 m / s 2 and v = 0 v = u + at = 30 10t t = 3 s

12 OpenStax-CNX module: m This means that this ball has actually traveled for 1 second (4 3 = 1 s) in the downward direction, when it is hit by the ball dropped from the tower! 4 Exercises Exercise 1 (Solution on p. 13.) A ball is thrown up in vertical direction with an initial speed of 40 m/s. Find acceleration of the ball at the highest point. Exercise 2 (Solution on p. 13.) A ball is released from a height of 45 m. Find the magnitude of average velocity during its motion till it reaches the ground. Exercise 3 (Solution on p. 13.) A ball is released from an elevator moving upward with an acceleration 3 m/s 2. What is the acceleration of the ball after it is released from the elevator? Exercise 4 (Solution on p. 13.) A ball is released from an elevator moving upward with an acceleration 5 m/s 2. What is the acceleration of the ball with respect to elevator after it is released from the elevator? Exercise 5 (Solution on p. 13.) A balloon ascends vertically with a constant speed for 5 second, when a pebble falls from it reaching the ground in 5 s. Find the speed of balloon. Exercise 6 (Solution on p. 14.) A balloon ascends vertically with a constant speed of 10 m/s. At a certain height, a pebble falls from it reaching the ground in 5 s. Find the height of ballon when pebble is released from the balloon. Exercise 7 (Solution on p. 14.) A ball is released from a top. Another ball is dropped from a point 15 m below the top, when the rst ball reaches a point 5 m below the top. Both balls reach the ground simultaneously. Determine the height of the top. Exercise 8 (Solution on p. 15.) One ball is dropped from the top at a height 60 m, when another ball is projected up in the same line of motion. Two balls hit each other 20 m below the top. Compare the speeds of the ball when they strike. note: See module titled Vertical motion under gravity (application) for more questions.

13 OpenStax-CNX module: m Solutions to Exercises in this Module The velocity of the ball at the highest point is zero. The only force on the ball is due to gravity. The accleration of ball all through out its motion is acceleration due to gravity g, which is directed downwards. The acceleration of the ball is constant and is not dependent on the state of motion - whether it is moving or is stationary. The average velocity is ratio of displacement and time. Here, displacement is given. We need to nd the time of travel. For the motion of ball, we consider the point of release as origin and upward direction as positive. y = ut at2 45 = 0Xt X 10Xt2 t 2 = 45 5 = 9 t = ±3 Neglecting negative time, t = 3 s. Magnitude of average velocity is : s v avg = 45 3 = 15 m/s On separation, ball acquires the velocity of elevator not its acceleration. Once it is released, the only force acting on it is that due to gravity. Hence, acceleration of the ball is same as that due to gravity. m a = 10 downward s 2 On separation, ball acquires the velocity of elevator not its acceleration. Once it is released, the only force acting on it is that due to gravity. Hence, acceleration of the ball is same as that due to gravity. The relative acceleration of the ball (considering downward direction as positive) : a rel = a ball a elevator m a rel = 10 ( 5) = 15 s 2 The velocity of balloon is constant and is a measured value. Let the ball moves up with a velocity u. At the time of release, the pebble acquires velocity of balloon. For the motion of pebble, we consider the point of release as origin and upward direction as positive. Here, Using equation for displacement : y = vt = ux5 = 5u; u = u; a = g; t = 5 s y = ut at2

14 OpenStax-CNX module: m u = ux X gx52 m 10u = 5X25u = 5 s The velocity of balloon is constant. Let the ball moves up with a velocity u. At the time of release, the pebble acquires velocity of balloon. For the motion of pebble, we consider the point of release as origin and upward direction as positive. Here, Using equation for displacement : y =?; u = 10 m/s; a = g; t = 5 s y = ut at2 y = 10X X gx52 y = 50 5X25 y = 75 Height of ballon when pebble is released from it, m H = 75 We compare motion of two balls under gravity, when second ball is dropped. At that moment, two balls are 10 m apart. The rst ball moves with certain velocity, whereas rst ball starts with zero velocity. Let us consider downward direction as positive. The velocity of the rst ball when it reaches 10 m below the top is : m v 2 = u 2 + 2ax v 2 = 0 + 2X10X5 v = 10 Let the balls take time t to reach the gorund. First ball travels 10 m more than second ball. Let 1 and 2 denote rst and second ball, then, m/s t X10t2 = 1 2 X10t2 10t = 10 t = 1 In this time, second ball travels a distance given by : s

15 OpenStax-CNX module: m y = 1 2 X10t2 = 5X12 = 5 m But, second ball is 15 m below the top. Hence, height of the top is = 20 m. For the motion of rst ball dropped from the top, let downward direction be positive : v 1 = u + at For the ball dropped from the top, v 1 = at = 10t x = ut at2 20 = 0Xt X 10Xt2 20 = 5t 2 t = ±2 s Neglecting negative value, t = 2s. Hence, velocity of the ball dropped from the top is : v 1 = 10t = 10X2 = 30 m/s For the motion of second ball projected from the bottom, let upward direction be positive : v 2 = u + at Clearly, we need to know u. For upward motion, v 2 = u 10X2 = u 20 x = ut at2 40 = ux X 10X22 40 = 2u 20 = 30m/s Thus, v 2 = = 10 v 1 v 2 = = 2 1 m/s

Vertical motion under gravity (application) *

OpenStax-CNX module: m14550 1 Vertical motion under gravity (application) * Sunil Kumar Singh This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License.0 Questions