This preview shows page 1. Sign up to view the full content.
Unformatted text preview: each ball. The data for the vertical motion are summarized as follows:
xA = xB v0 x,A tA = v0 x,BtB or or yDirection Data, Ball A yA ay,A −1.2 m vy,A −9.80 m/s2 yDirection Data, Ball B v0y,A tA yB ay,B 0 m/s ? −1.5 m −9.80 m/s2 vy,B v0y,B tB 0 m/s ? Note that the initial velocity components v0y are zero, because the balls are bunted
horizontally. With these data, Equation 3.5b gives
y = v0 y t + a y t 2 = ( 0 m/s ) t + a y t 2
1
2 1
2 or t= 2y
ay (2) SOLUTION Using Equation (2) for each ball and substituting the expressions for tA and tB
into Equation (1) gives
2 yA / a y ,A
t v0 x,B = v0 x,A A = v0 x,A
t 2 yB / a y ,B B = v0 x,A yA
1.2 m
= (1.9 m/s )
= 1.7 m/s
yB
1.5 m Note that the accelerations ay,A and ay,B both equal the acceleration due to gravity and are
eliminated algebraically from the calculation. 73. REASONING The drawing shows the
trajectory of the ball, along with its
initial speed v0 and vertical displacement
y. The angle that the initial velocity of
the ball makes with the ground is 35.0°.
The known data are shown in the table
below: +y
+x v0
35.0° y Chapter 3 Problems 151 yDirection Data
y ay +5.50 m −9.80 m/s2 v0y t +(46.0 m/s) sin 35.0° = +26.4 m/s ? vy Since three of the kinematic variables are known, we will employ the appropriate equation
of kinematics to determine the time of flight.
SOLUTION Equation 3.5b (y =v 0y t + 1 ayt 2
2 ) relates the time t to the three known variables. The terms in this equation can be rearranged to as to place it in a standard
quadratic form: 1 a y t 2 + v0 y t − y = 0 . The solution of this quadratic equation is
2 t= t= ( a )(− y)
2( a ) 2
−v0 y ± v0 y − 4
1
2 1
2 y y − ( +26.4 m/s ) ± ( +26.4 m/s ) 2 − 4 ( 1 ) ( −9.80 m/s 2 ) ( −5.50 m )
2 −9.80 m/s 2 = 0.217 s or 5.17 s The first solution (t = 0.217 s) corresponds to the situation where the ball is moving upward
and has a displacement of y = +5.50 m. The second solution represents the later time when
the ball is moving downward and its displacement is also y = +5.50 m (see the drawing).
This is the solution we seek, so t = 5.17 s .
______________________________________________________________________________
74. REASONING There are three velocities involved:
v CR = the initial velocity of the Car relative to the Road
v C T = the velocity of the Car relative to the first Truck
v TR = the velocity of the first Truck relative to the Road (speed = 11 m/s) Ordering these vectors by their subscripts in the manner discussed in Section 3.4 of the text,
we see that vCR = vCT + vTR. The velocity v TR is known, and we will use the equations of
kinematics to find vCT.
The two trucks have the same velocity relative to the road, and thus have zero velocity
relative to one another. Therefore, we will analyze the sports car’s jump across the 15m gap
as if both trucks were stationary. Neglecting air resistance, we treat the car as a projectile
and find the initial speed v0 it must have, relative to the trucks, in order to reach the flat
trailer. (This speed is, in fact, the speed vCT that we need to find.) We will use the relation 152 KINEMATICS IN TWO DIMENSIONS x = v0 x t + 1 a x t 2 (Equation 3.5a) to find the elapsed time. With ax = 0 m/s2, Equation 3.5a
2 ( ) becomes x = v0 xt + 1 0 m/s 2 t 2 = v0 x t . Thus the car is in the air for t seconds, where
2
t= x
v0 x x
v0 cos θ = (1) In Equation (1), x = 15 m and θ = 16º, but v0 and t are not known. The heights of the trailer
and the ramp are the same, so the car’s vertical displacement is zero. To use this fact, we
turn to y = v0 yt + 1 ayt 2
2 (Equation 3.5b) and substitute both y=0m and v0y = v0 sin θ:
0 = v0 y t + 1 a y t 2
2 or v0 y = − 1 a y t
2 v0 sin θ = − 1 a y t
2 or (2) To eliminate the time, we substitute t from Equation (1) into Equation (2) and find that x
1
v0 sin θ = − 2 a y v0 cos θ or 2
v0 = −a y x
2 cos θ sin θ or v0 = −a y x
2 cos θ sin θ (3) SOLUTION Taking up as the positive direction, we use Equation (3) to calculate the speed
vCT = v0 of the car relative to the trucks: vCT = v0 = −a y x
2 cos θ sin θ = ( ) − −9.80 m/s 2 (15 m )
2 cos16° sin16o = 17 m/s Since the ramp alters the direction of the car’s velocity but not its magnitude, the initial
jump speed v0 is also the magnitude of the car’s velocity vCT relative to the truck before the
car reaches the ramp: v0 = vCT. The sports car must overtake the truck at a speed of at least
17 m/s relative to the truck, so that the car’s minimum required speed vCR relative to the
road is
vCR = vCT + vTR = 17 m/s + 11 m/s = 28 m/s Chapter 3 Problems 75. SSM 153 REASONING The angle θ can be found from 2400 m x θ = tan −1 (1) where x is the horizontal displacement of the flare. Since ax = 0 m/s2, it follows that x = (v0 cos 30.0°)t . The flight time t is determined by the vertical motion. In particular, the
time t can be found from Equation 3.5b. Once the time is known, x can be calculated. SOLUTION From Equation 3.5b, assuming upward is the positive direction...
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