Physics Solution Manual for 1100 and 2101

# 9 v0 v 2 2ay 27 ms 2 2980 ms 2 15 m 209

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Unformatted text preview: the pellet would have gone if the gun had been fired straight upward, provided that we can determine the initial speed imparted to the pellet by the gun. This initial speed can be found by applying Equation 2.9 to the downward motion of the pellet described in the problem statement. SOLUTION If we assume that upward is the positive direction, the initial speed of the pellet is, from Equation 2.9, v0 = v 2 − 2ay = ( – 27 m/s) 2 − 2(–9.80 m/s 2 )(– 15 m) = 20.9 m/s 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 −v0 −(20.9 m/s) 2 = 22 m 2a 2(–9.80 m/s2 ) ______________________________________________________________________________ y= = 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 ( y = v0t + 1 at 2 ) with v0 = 0 m/s. 2 Assuming upward to be the positive direction, we find 2y 2(–6.0 m) = = 1.1 s a –9.80 m/s 2 ______________________________________________________________________________ t= 54. REASONING (v 2 2 = v0 ) The initial speed of the ball can be determined from Equation 2.9 + 2ay . 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 v0 = v 2 − 2ay = ( 0 m/s )2 − 2(–9.80 m/s 2 )(12.0 m) = +15.3 m/s When the speed of the ball has decreased to one half of its initial value, v = 1 v0 , and 2 Equation 2.9 gives Chapter 2 Problems y= 2 v 2 − v0 2a = 2 ( 1 v0 ) 2 − v0 2 2a 73 2 v0 1 (+15.3 m/s)2 1 = − 1 = − 1 = 8.96 m 2a 4 2(–9.80 m/s 2 ) 4 55. REASONING The displacement y of the diver is equal to her average velocity v multiplied by the time t, or y = v t . Since the diver has a constant acceleration (the acceleration due to gravity), her average velocity is equal to v = 1 2 ( v0 + v ) , where v0 and v are, respectively, the initial and final velocities. Thus, according to Equation 2.7, the displacement of the diver is (2.7) y = 1 ( v0 + v ) t 2 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 − v0, divided by the elapsed time t: a = ( v − v0 ) / t . Solving this equation for the initial velocity v0 yields v0 = v − at Substituting this relation for v0 into Equation 2.7, we obtain y= 1 2 ( v0 + v ) t = 1 ( v − at + v ) t = vt − 1 at 2 2 2 2 SOLUTION The diver’s acceleration is that due to gravity, or a = −9.80 m/s . 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 y = vt − 1 at 2 = ( −10.1 m/s )(1.20 s ) − 1 ( −9.80 m/s2 ) (1.20 s ) = −5.06 m 2 2 2 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 divided into three intervals: from release (h1 = 9.50 m) to impact, from impact to the second highest point (h2 = 5.70 m), and from the second highest point to h3 = 1.20 m above the pavement. For each of the intervals, the acceleration is that due to gravity. For the first and last interval, the ball’s initial velocity is zero, so the time to fall a given distance can be ( found from Equation 2.8 y = v0t + 1 at 2 2 ). 74 KINEMATICS IN ONE DIMENSION The second interval begins at the pavement and ends at h2, so the initial velocity isn’t zero. However, the symmetry of free-fall motion is such that it takes the ball as much time to rise from the ground to a maximum height h2 as it would take for a ball dropped from h2 to fall to the pavement, so we can again use Equation 2.8 to find the duration of the second interval. SOLUTION Taking upward as the positive direction, we have a = −9.80 m/s2 for the acceleration in each of the three intervals. Furthermore, the initial velocity for each of the intervals is v0 = 0 m/s. Remember, we are using symmetry to treat the second interval as if the ball were dropped from rest at a height of 5.70 m and fell to the pavement. Using ( Equation 2.8 y = v0t + 1 at 2 2 ) , with v =...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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