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If the building is 30.0 m high, how long does it take for the ball to hit the ground? What velocity is the ball traveling at when it hits the ground? Answers 4.44 s, -28.5m/s Both images have background info to help solve the practice problem

If the building is 30.0 m high, how long does it take for the ball to hit the ground? What velocity is the ball traveling at when it hits the ground? Answers 4.44 s, -28.5m/s Both images have background info to help solve the practice problem

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If the building is 30.0 m high, how long does it take for the ball to hit the ground? What velocity is the ball traveling at when it hits the ground? Answers 4.44 s, -28.5m/s Both images have background info to help solve the practice problem
Unic Hashes. The flashes occur at equal time intervals, so this photograph records the positions of the ball at equal time intervals. Therefore, the average velocity of the ball between successive flashes is proportional to the distance between the corresponding images of the ball in the photograph. The increasing distances between images show that the ball's velocity increases continuously; in other words, the ball ac- celerates. Careful measurement shows that the change in velocity is the same in each time interval; thus the acceleration is constant. ► FIGURE 2.26 Strobe photo of a freely falling ball. EXAMPLE 2.11 A ball on the roof Now let's examine the more complex case of a ball being thrown vertically upward. Even though the ball is initially moving upward, it is, nevertheless, in free fall and we can use Equations 2.12, 2.13, and 2.14 to analyze its motion. Suppose you throw a ball vertically upward from the flat roof of a tall building. The ball leaves your hand at a point even with the roof railing, with an upward velocity of 15.0 m/s. On its way back down, it just misses the railing. Find (a) the position and velocity of the ball 1.00 s and 4.00 s after it leaves your hand; (b) the velocity of the ball when it is 5.00 m above the railing; and (c) the maximum height reached and the time at which it is reached. Ignore the effects of the air. OSMO O
Transcribed Image Text:Unic Hashes. The flashes occur at equal time intervals, so this photograph records the positions of the ball at equal time intervals. Therefore, the average velocity of the ball between successive flashes is proportional to the distance between the corresponding images of the ball in the photograph. The increasing distances between images show that the ball's velocity increases continuously; in other words, the ball ac- celerates. Careful measurement shows that the change in velocity is the same in each time interval; thus the acceleration is constant. ► FIGURE 2.26 Strobe photo of a freely falling ball. EXAMPLE 2.11 A ball on the roof Now let's examine the more complex case of a ball being thrown vertically upward. Even though the ball is initially moving upward, it is, nevertheless, in free fall and we can use Equations 2.12, 2.13, and 2.14 to analyze its motion. Suppose you throw a ball vertically upward from the flat roof of a tall building. The ball leaves your hand at a point even with the roof railing, with an upward velocity of 15.0 m/s. On its way back down, it just misses the railing. Find (a) the position and velocity of the ball 1.00 s and 4.00 s after it leaves your hand; (b) the velocity of the ball when it is 5.00 m above the railing; and (c) the maximum height reached and the time at which it is reached. Ignore the effects of the air. OSMO O
50 CHAPTER 2 Motion Along a Straight Line SOLUTION SET UP As shown in Figure 2.27, we place the origin at the level of the roof railing, where the ball leaves your hand, and we take the posi- tive direction to be upward. Here's what we know: The initial position % is zero, the initial velocity voy (y component) is +15.0 m/s, and the acceleration (y component) is a, = -9.80 m/s². The ball actually moves straight up and then straight down; we show. a U-shaped path for clarity. YU₂ = ? t = 1.00 s U₁ = ? t = ? t = 0, Voy A FIGURE 2.27 15.0 m/s 1 U₂ = 0 1 = ? --- ---y = 5.00 m t = ? Uy = ? t = 4.00 s U₁ = ? y = 0 ay = -8 = -9.80 m/s² y = ? N o rather that the dat sommin SOLVE What equations do we have to work with? The velocity u, at any time ris Vy= Voy+ ayt = 15.0 m/s + (-9.80 m/s²)t. The position y at any time t is y = voyt + a,t² = (15.0 m/s)t + (-9.80 m/s²);². The velocity u, at any position y is given by 2 v² = voy² + 2a,(y-yo) = (15.0 m/s)2 + 2(-9.80 m/s²)(y-0). Part (a): When = 1.00 s, the first two equations give y = +10.1m, 1 = Uy +5.20 m/s. The ball is 10.1 m above the origin (y is positive), and it has an upward velocity (v is positive) of 5.20 m/s (less than the initial velocity of 15.0 m/s, as expected). When t = 4.00 s, the same equations give y = -18.4 m, u, = -24.2 m/s. Thus, at time t = 4.00 s, the ball has passed its highest point and is 18.4 m below the origin (y is negative). It has a downward velocity (u, is negative) with magnitude 24.2 m/s (greater than the initial velocity, as we should expect). Note that, to get these results, we don't need to find the highest point the ball reaches or the time at which it is reached. The equations of motion give the position and velocity at any time, whether the ball is on the way up or on the way down. Part (b): When the ball is 5.00 m above the origin, y = +5.00 m. Now we use our third equation to find the velocity u, at this point: v² = (15.0 m/s)2 + 2(-9.80 m/s2) (5.00 m) = 127 m²/s², Uy = ±11.3 m/s. We get two values of Uy, one positive and one negative. That is, the ball passes the point y = +5.00 m twice, once on the way up and again on the way down. The velocity on the way up is +11.3 m/s, and on the way down it is -11.3 m/s. Part (c): At the highest point, the ball's velocity is momentarily zero (vy=0); it has been going up (positive u,) and is about to start going down (negative uy). From our third equation, we have 0 = (15.0 m/s)² (19.6 m/s²)y, and the maximum height (where uy = 0) is y = 11.5 m. We can now find the time when the ball reaches its highest point from Equation 2.12, setting u, = 0: 0 15.0 m/s + (-9.80 m/s²)r, t = 1.53 s. Alternative Solution: Alternatively, to find the maximum height, we may ask first when the maximum height is reached. That is, at what value of t is v, = 0? As we just found, uy = 0 when t = 1.53 s. Substi- tuting this value of t back into the equation for y, we find that y = (15 m/s) (1.53 s) + (-9.8 m/s²) (1.53 s)² = 11.5 m. REFLECT Although at the highest point the velocity is momentarily zero, the acceleration at that point is still -9.80 m/s². The ball stops for an instant, but its velocity is continuously changing, from positive values through zero to negative values. The acceleration is constant throughout; it is not zero when u, = 0! Figure 2.28 shows graphs of position and velocity as functions of time for this problem. Note that the graph of uy versus t has a constant negative slope. Thus, the acceleration is negative on the way up, at the highest point, and on the way down. Practice Problem: If the building is 30.0 m high, how long does it take for the ball to hit the ground? What velocity is the ball traveling at when it hits the ground? Answers: 4.44 s, -28.5 m/s. y (m) 15 10 5 0 -5- -10- -15 -20 F (a) A FIGURE 2.28 (a) Position and (b) velocity as functions of time for a ball thrown upward with an initial velocity of 15 m/s. 1 1 2 3 1(s) vy (m/s) 15 10 .5 0 -5- -10- -15 -20 -25 (b) 1 1 The acceleration is constant (straight line) and negative (negative slope). L 2 L 3 4 (s); <
Transcribed Image Text:50 CHAPTER 2 Motion Along a Straight Line SOLUTION SET UP As shown in Figure 2.27, we place the origin at the level of the roof railing, where the ball leaves your hand, and we take the posi- tive direction to be upward. Here's what we know: The initial position % is zero, the initial velocity voy (y component) is +15.0 m/s, and the acceleration (y component) is a, = -9.80 m/s². The ball actually moves straight up and then straight down; we show. a U-shaped path for clarity. YU₂ = ? t = 1.00 s U₁ = ? t = ? t = 0, Voy A FIGURE 2.27 15.0 m/s 1 U₂ = 0 1 = ? --- ---y = 5.00 m t = ? Uy = ? t = 4.00 s U₁ = ? y = 0 ay = -8 = -9.80 m/s² y = ? N o rather that the dat sommin SOLVE What equations do we have to work with? The velocity u, at any time ris Vy= Voy+ ayt = 15.0 m/s + (-9.80 m/s²)t. The position y at any time t is y = voyt + a,t² = (15.0 m/s)t + (-9.80 m/s²);². The velocity u, at any position y is given by 2 v² = voy² + 2a,(y-yo) = (15.0 m/s)2 + 2(-9.80 m/s²)(y-0). Part (a): When = 1.00 s, the first two equations give y = +10.1m, 1 = Uy +5.20 m/s. The ball is 10.1 m above the origin (y is positive), and it has an upward velocity (v is positive) of 5.20 m/s (less than the initial velocity of 15.0 m/s, as expected). When t = 4.00 s, the same equations give y = -18.4 m, u, = -24.2 m/s. Thus, at time t = 4.00 s, the ball has passed its highest point and is 18.4 m below the origin (y is negative). It has a downward velocity (u, is negative) with magnitude 24.2 m/s (greater than the initial velocity, as we should expect). Note that, to get these results, we don't need to find the highest point the ball reaches or the time at which it is reached. The equations of motion give the position and velocity at any time, whether the ball is on the way up or on the way down. Part (b): When the ball is 5.00 m above the origin, y = +5.00 m. Now we use our third equation to find the velocity u, at this point: v² = (15.0 m/s)2 + 2(-9.80 m/s2) (5.00 m) = 127 m²/s², Uy = ±11.3 m/s. We get two values of Uy, one positive and one negative. That is, the ball passes the point y = +5.00 m twice, once on the way up and again on the way down. The velocity on the way up is +11.3 m/s, and on the way down it is -11.3 m/s. Part (c): At the highest point, the ball's velocity is momentarily zero (vy=0); it has been going up (positive u,) and is about to start going down (negative uy). From our third equation, we have 0 = (15.0 m/s)² (19.6 m/s²)y, and the maximum height (where uy = 0) is y = 11.5 m. We can now find the time when the ball reaches its highest point from Equation 2.12, setting u, = 0: 0 15.0 m/s + (-9.80 m/s²)r, t = 1.53 s. Alternative Solution: Alternatively, to find the maximum height, we may ask first when the maximum height is reached. That is, at what value of t is v, = 0? As we just found, uy = 0 when t = 1.53 s. Substi- tuting this value of t back into the equation for y, we find that y = (15 m/s) (1.53 s) + (-9.8 m/s²) (1.53 s)² = 11.5 m. REFLECT Although at the highest point the velocity is momentarily zero, the acceleration at that point is still -9.80 m/s². The ball stops for an instant, but its velocity is continuously changing, from positive values through zero to negative values. The acceleration is constant throughout; it is not zero when u, = 0! Figure 2.28 shows graphs of position and velocity as functions of time for this problem. Note that the graph of uy versus t has a constant negative slope. Thus, the acceleration is negative on the way up, at the highest point, and on the way down. Practice Problem: If the building is 30.0 m high, how long does it take for the ball to hit the ground? What velocity is the ball traveling at when it hits the ground? Answers: 4.44 s, -28.5 m/s. y (m) 15 10 5 0 -5- -10- -15 -20 F (a) A FIGURE 2.28 (a) Position and (b) velocity as functions of time for a ball thrown upward with an initial velocity of 15 m/s. 1 1 2 3 1(s) vy (m/s) 15 10 .5 0 -5- -10- -15 -20 -25 (b) 1 1 The acceleration is constant (straight line) and negative (negative slope). L 2 L 3 4 (s); <
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