This preview shows page 1. Sign up to view the full content.
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 1.20s 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 freefall 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 freefall 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 =...
View
Full
Document
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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details