{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics Solution Manual for 1100 and 2101

The two paths differ by the extra distance covered by

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 35 m/s. The final velocity is 2 v = v0 + at = 35 m/s + ( 9.8 m/s )(5.0 s) = − 14 m/s (2.4) Instantaneous speed is the magnitude of the instantaneous velocity, so we drop the minus sign and find that Speed = 14 m/s 45. REASONING AND SOLUTION The figure at the right shows the paths taken by the pellets fired from gun A and gun B. The two paths differ by the extra distance covered by the pellet from gun A as it rises and falls back to the edge of the cliff. When it falls back to the edge of the cliff, the pellet from gun A will have the same speed as the pellet fired from gun B, as Conceptual Example 15 discusses. Therefore, the flight time of pellet A will be greater than that of B by the amount of time that it takes for pellet A to cover the extra distance. A ____ B The time required for pellet A to return to the cliff edge after being fired can be found from Equation 2.4: v = v0 + a t. If "up" is taken as the positive direction then v0 = +30.0 m/s and v = –30.0 m/s. Solving Equation 2.4 for t gives t= v − v0 a = (−30.0 m/s) − (+30.0 m/s) = −9.80 m/s2 6.12 s Notice that this result is independent of the height of the cliff. ______________________________________________________________________________ 46. REASONING Assuming that air resistance can be neglected, the acceleration is the same for both the upward and downward parts, namely −9.80 m/s2 (upward is the positive direction). Moreover, the displacement is y = 0 m, since the final and initial positions of the ball are the 70 KINEMATICS IN ONE DIMENSION same. The time is given as t = 8.0 s. ( y = v0 t + 1 a t2 2 Therefore, we may use Equation 2.8 ) to find the initial velocity v0 of the ball. ( ) SOLUTION Solving Equation 2.8 y = v0 t + 1 a t 2 for the initial velocity v0 gives 2 v0 = y − 1 a t2 2 ( ) 0 m − 1 −9.80 m/s2 ( 8.0 s ) 2 2 = = +39 m/s t 8.0 s ______________________________________________________________________________ 47. REASONING AND SOLUTION In a time t the card will undergo a vertical displacement y given by y = 1 at2 2 where a = –9.80 m/s2. When t = 60.0 ms = 6.0 × 10–2 s, the displacement of the card is 0.018 m, and the distance is the magnitude of this value or d1 = 0.018 m . Similarly, when t = 120 ms, d 2 = 0.071 m , and when t = 180 ms, d3 = 0.16 m . ______________________________________________________________________________ 48. REASONING The minimum time that a player must wait before touching the basketball is the time required for the ball to reach its maximum height. The initial and final velocities are known, as well as the acceleration due to gravity, so Equation 2.4 ( v = v0 + a t ) can be used to find the time. SOLUTION Solving Equation 2.4 for the time yields v − v0 0 m / s − 4.6 m / s = 0.47 s a −9.8 m / s 2 ______________________________________________________________________________ t= 49. = SSM REASONING AND SOLUTION Equation 2.8 can be used to determine the displacement that the ball covers as it falls halfway to the ground. Since the ball falls from rest, its initial velocity is zero. Taking down to be the negative direction, we have y = v0t + 1 at 2 = 1 at 2 = 1 (–9.80 m/s 2 )(1.2 s) 2 = –7.1 m 2 2 2 Chapter 2 Problems 71 In falling all the way to the ground, the ball has a displacement of y = –14.2 m . Solving Equation 2.8 with this displacement then yields the time 2y 2(–14.2 m) = = 1.7 s a –9.80 m/s 2 ______________________________________________________________________________ t= 50. REASONING The initial velocity of the compass is +2.50 m/s. The initial position of the compass is 3.00 m and its final position is 0 m when it strikes the ground. The displacement of the compass is the final position minus the initial position, or y = –3.00 m. As the compass falls to the ground, its acceleration is the acceleration due to gravity, a = –9.80 m/s2. Equation 2.8 y = v0t + 1 a t 2 can be used to find how much time elapses ( 2 ) before the compass hits the ground. SOLUTION Starting with Equation 2.8, we use the quadratic equation to find the elapsed time. t= ( 1 a ) ( − y ) = − ( 2.50 m/s ) ± ( 2.50 m/s )2 − 4 ( −4.90 m/s2 ) − ( −3.00 m ) 2 1a 2 2( 2 ) 2 ( −4.90 m/s ) 2 −v0 ± v0 − 4 There are two solutions to this quadratic equation, t1 = 1.08 s and t2 = –0.568 s. The second solution, being a negative time, is discarded. ______________________________________________________________________________ 51. REASONING AND SOLUTION a. 2 v 2 = v0 + 2ay ( ) v = ± (1.8 m/s ) + 2 –9.80 m/s 2 ( –3.0 m ) = ±7.9 m/s 2 The minus is chosen, since the diver is now moving down. Hence, v = −7.9 m/s . b. The diver's velocity is zero at his highest point. The position of the diver relative to the board is y=– 2 v0 2a =– (1.8 m/s )2 ( 2 –9.80 m/s 2 ) = 0.17 m The position above the water is 3.0 m + 0.17 m = 3.2 m . ______________________________________________________________________________ 72 KINEMATICS IN ONE DIMENSION ( ) 2 Equation 2.9 v 2 = v0 + 2ay can be used to find out how far above the 52. REASONING cliff's edge...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern