SOLUTION If we assume that upward is the positive direction the initial speed

Solution if we assume that upward is the positive

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SOLUTION If we assume that upward is the positive direction, the initial speed of the pellet is, from Equation 2.9, 2 2 2 0 2 ( – 27 m/s) 2(–9.80 m/s )(– 15 m) 20.9 m/s v v ay = - = - =
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Chapter 2 Problems 73 Equation 2.9 can again be used to find the maximum height of the pellet if it were fired straight up. At its maximum height, v = 0 m/s, and Equation 2.9 gives 2 2 0 2 (20.9 m/s) 22 m 2 2(–9.80 m/s ) v y a - - = = = _____________________________________________________________________________ _ 53. SSM REASONING AND SOLUTION Since the balloon is released from rest, its initial velocity is zero. The time required to fall through a vertical displacement y can be found from Equation 2.8 ( 29 2 1 0 2 y v t at = + with 0 0 v = m/s. Assuming upward to be the positive direction, we find 2 2 2(–6.0 m) 1.1 s –9.80 m/s y t a = = = _____________________________________________________________________________ _ 54. REASONING The initial speed of the ball can be determined from Equation 2.9 ( 29 2 2 0 2 v v ay = + . Once the initial speed of the ball is known, Equation 2.9 can be used a second time to determine the height above the launch point when the speed of the ball has decreased to one half of its initial value. SOLUTION When the ball has reached its maximum height, its velocity is zero. If we take upward as the positive direction, we have from Equation 2.9 ( 29 2 2 2 0 2 0 m/s 2(–9.80 m/s )(12.0 m) 15.3 m/s v v ay = - = - = + When the speed of the ball has decreased to one half of its initial value, 1 0 2 v v = , and Equation 2.9 gives 2 2 2 2 2 1 2 0 0 0 0 2 2 ( ) 1 (+15.3 m/s) 1 1 1 8.96 m 2 2 2 4 4 2(–9.80 m/s ) v v v v v y a a a - - = = = - = - = ÷ ÷ 55. REASONING The displacement y of the diver is equal to her average velocity v multiplied by the time t , or y vt = . Since the diver has a constant acceleration (the acceleration due to gravity), her average velocity is equal to ( 29 1 0 2 v v v = + , where v 0 and v are, respectively, the initial and final velocities. Thus, according to Equation 2.7, the displacement of the diver is ( 29 1 0 2 y v v t = + (2.7)
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74 KINEMATICS IN ONE DIMENSION The final velocity and the time in this expression are known, but the initial velocity is not. To determine her velocity at the beginning of the 1.20-s period (her initial velocity), we turn to her acceleration. The acceleration is defined by Equation 2.4 as the change in her velocity, v - v 0 , divided by the elapsed time t : ( 29 0 / a v v t = - . Solving this equation for the initial velocity v 0 yields 0 v v at = - Substituting this relation for v 0 into Equation 2.7, we obtain ( 29 ( 29 2 1 1 1 0 2 2 2 y v v t v at v t vt at = + = - + = - SOLUTION The diver’s acceleration is that due to gravity, or a = - 9.80 m/s 2 . The acceleration is negative because it points downward, and this direction is the negative direction. The displacement of the diver during the last 1.20 s of the dive is ( 29 ( 29 ( 29 ( 29 2 2 2 1 1 2 2 10.1 m/s 1.20 s 9.80 m/s 1.20 s 5.06 m y vt at = - = - - - = - The displacement of the diver is negative because she is moving downward. _____________________________________________________________________________ _ 56. REASONING The ball is initially in free fall, then collides with the pavement and rebounds, which puts it into free fall again, until caught by the boy. We don’t have enough information to analyze its collision with the pavement, but we’re only asked to calculate the time it spends in the air, undergoing free-fall motion. The motion can be conveniently
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  • Spring '11
  • rollino
  • Physics, Velocity, m/s

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