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Unformatted text preview: v0y t 0 m/s tB With these data, Equation 3.5b gives
y = v0 y t + a y t 2 = ( 0 m/s ) t + a y t 2 = a y t 2
1
2 1
2 1
2 (1) SOLUTION Applying Equation (1) to both shots, we find that yB
yA = −HB −H A = 1
a t2
2 yB
1
a t2
2 yA or HB
HA = 2
tB (2) 2
tA To use this result to calculate the ratio HB/HA, it is necessary to determine the times tA and
tB. To do this, we consider the horizontal part of the motion and note that there is no
acceleration in the horizontal direction. Therefore, the horizontal component v0x of the
bullet’s initial velocity remains unchanged throughout the motion, and the horizontal
component x of the displacement is simply v0x times the time t during which the motion
occurs. We have that xA = v0 x tA and xB = v0 xtB or Substituting these results into Equation (2) gives tA = xA
v0 x and tB = xB
v0 x (3) Chapter 3 Problems 2
x H B tB ( xB / v0 x )
=2=
= B H A tA ( x / v )2 xA A
0x
2 121 2 It is given that xB = 2xA, so that we find HB
HA 37. x
= B
x
A 2 2 xA = xA 2 =4 SSM REASONING
a. The drawing shows the initial velocity v0 of the
package when it is released. The initial speed of the
package is 97.5 m/s. The component of its
displacement along the ground is labeled as x. The
data for the x direction are indicated in the data
table below. +y v0 +x 50.0° y x
xDirection Data x ax ? 0 m/s2 vx v0x t +(97.5 m/s) cos 50.0° = +62.7 m/s Since only two variables are known, it is not possible to determine x from the data in this
table. A value for a third variable is needed. We know that the time of flight t is the same for
both the x and y motions, so let’s now look at the data in the y direction.
yDirection Data y ay −732 m −9.80 m/s vy
2 v0y t +(97.5 m/s) sin 50.0° = +74.7 m/s ? Note that the displacement y of the package points from its initial position toward the
ground, so its value is negative, i.e., y = −732 m. The data in this table, along with the
appropriate equation of kinematics, can be used to find the time of flight t. This value for t
can, in turn, be used in conjunction with the xdirection data to determine x. 122 KINEMATICS IN TWO DIMENSIONS
+y b. +x The drawing at the right shows the velocity of the package
just before impact. The angle that the velocity makes with
respect to the ground can be found from the inverse ( ) tangent function as θ = tan −1 v y / vx . Once the time has
been found in part (a), the values of vy and vx can be
determined from the data in the tables and the appropriate
equations of kinematics. vx
θ vy v
SOLUTION
a. To determine the time that the package is in the air, we will use Equation 3.5b (y =v 0y ) t + 1 a y t 2 and the data in the ydirection data table. Solving this quadratic equation
2 for the time yields t= t= ( a )(− y)
2( a ) 2
−v0 y ± v0 y − 4
1
2 − ( 74.7 m/s ) ± 1
2 y y ( 74.7 m/s ) − 4 ( 1 ) ( −9.80 m/s 2 ) ( 732 m )
2
=
1
2 ( 2 ) ( −9.80 m/s 2 )
2 − 6.78 s and 22.0 s We discard the first solution, since it is a negative value and, hence, unrealistic. The
displacement x can be found using t = 22.0 s, the data in the xdirection data table, and
Equation 3.5a:
x = v0 x t + 1 ax t 2 = ( +62.7 m/s )( 22.0 s ) +
2 ( 0 m/s2444) = +1380 m
) ( 22.0 s3
1444
1
2 2 2 =0 b. The angle θ that the velocity of the package makes with respect to the ground is given by ( ) θ = tan −1 v y / vx . Since there is no acceleration in the x direction (ax = 0 m/s2), vx is the
same as v0x, so that vx = v0x = +62.7 m/s. Equation 3.3b can be employed with the
ydirection data to find vy :
v y = v0 y + a y t = +74.7 m/s + ( −9.80 m/s 2 ) ( 22.0 s ) = −141 m/s
Therefore, vy −1 −141 m/s = tan = −66.0° vx +62.7 m/s θ = tan −1 where the minus sign indicates that the angle is 66.0° below the horizontal .
______________________________________________________________________________ Chapter 3 Problems 123 38. REASONING We can obtain an expression for the car’s initial velocity v0 by starting with
the relation x = v0 x t + 1 a x t 2 (Equation 3.5a). Here x is the horizontal component of the car’s
2
displacement, v0x is the horizontal component of the car’s initial velocity
(v0x = v0 for a horizontal launch), and ax is its acceleration in the horizontal direction
(ax = 0 m/s2 for projectile motion). Solving for the initial velocity, we find that ( ) x = v0t + 1 0 m/s2 t 2 = v0t
2 or v0 = x
t (1) The time t in Equation (1) is known, but x is not. To find x, we note that at t = 1.1 s, the car’s
displacement has a magnitude of ∆r = 7.0 m. The displacement ∆r of the car has a
horizontal component x and a vertical component y. The magnitude of the displacement is
related to x and y by the Pythagorean theorem:
(2)
( ∆r )2 = x 2 + y 2
Solving Equation (2) for x and substituting the result into Equation (1) gives
v0 = x
=
t ( ∆r )2 − y 2 (3) t To determine y, we turn to the relation y = v0 yt + 1 a yt 2 (Equation 3.5b). Setting
2
v0y = 0 m/s (again, for a horizontal launch), we find that Equation (3) becomes y = ( 0...
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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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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