Unformatted text preview: , we have y = –(v0 sin 30.0°) t + 1 a y t 2
2
which can be rearranged to give the following equation that is quadratic in t:
1 a t2
2y − (v0 sin 30.0°) t − y = 0 Using y = –2400 m and a y = −9.80 m/s 2 and suppressing the units, we obtain the quadratic
equation
4.9t 2 + 120t − 2400 = 0 Using the quadratic formula, we obtain t = 13 s. Therefore, we find that
x = ( v 0 cos 30 .0 ° ) t = ( 240 m / s ) (cos 30.0 ° ) ( 13 s) = 2700 m Equation (1) then gives 2400 m = 42° 2700 m θ = tan −1 76. REASONING The relative velocities in this problem are:
vPW = velocity of the Passenger relative to the Water
vPB = velocity of the Passenger relative to the Boat (2.50 m/s, due east)
vBW = velocity of the Boat relative to the Water (5.50 m/s, at 38.0° north of east) The velocities are shown in the drawing are related by the subscripting method discussed in
Section 3.4:
vPW = vPB + vBW 154 KINEMATICS IN TWO DIMENSIONS +y (North) vPW vPW +x (East) 5.50 m/s vBW
θ θ vPB 2.50 m/s 38.0° We will determine the magnitude and direction of vPW from the equation above by using the
method of scalar components.
SOLUTION The table below lists the scalar components of the three vectors.
Vector x Component y Component vPB +2.50 m/s 0 m/s vBW
vPW = vPB + vBW +(5.50 m/s) cos 38.0° = +4.33 m/s +(5.50 m/s) sin 38.0° = +3.39 m/s
+2.50 m/s + 4.33 m/s = +6.83 m/s 0 m/s + 3.39 m/s = +3.39 m/s The magnitude of vPW can be found by applying the Pythagorean theorem to the x and y
components:
vPW = ( 6.83 m/s ) + ( 3.39 m/s )
2 2 = 7.63 m/s The angle θ (see the drawings) that vPW makes with due east is +3.39 m/s = 26.4° north of east +6.83 m/s ______________________________________________________________________________ θ = tan −1 77. REASONING The drawings show the initial and final velocities of the ski jumper and their
2
2
scalar components. The initial speed of the ski jumper is given by v0 = v0 x + v0 y , and the v angle that the initial velocity makes with the horizontal is θ = tan −1 0 y . The scalar
v 0x components v0x and v0y can be determined by using the equations of kinematics and the data
in the following tables. (The +x direction is in the direction of the horizontal displacement
of the skier, and the +y direction is “up.”) Chapter 3 Problems v0 155 vx v0y 43.0° θ v v0x
Initial velocity vy Final velocity xDirection Data
x ax vx v0x +51.0 m 0 m/s2 +(23.0 m/s) cos 43.0° = +16.8 m/s ? t yDirection Data
y ay
−9.80 m/s vy
2 v0y −(23.0 m/s) sin 43.0° = −15.7 m/s t ? Since there is no acceleration in the x direction (ax = 0 m/s2), v0x is the same as vx, so we
have that v0x = vx = +16.8 m/s. The time that the skier is in the air can be found from the
xdirection data, since three of the variables are known. With the value for the time and the
ydirection data, the y component of the initial velocity can be determined.
2 SOLUTION Since ax = 0 m/s , the time can be determined from Equation 3.5a as
x
+51.0 m
=
= 3.04 s . The value for v0y can now be found by using Equation 3.3b
t=
v0 x +16.8 m/s
with this value of the time and the ydirection data:
v0 y = v y − a y t = −15.7 m/s − ( −9.80 m/s 2 ) ( 3.04 s ) = +14.1 m/s The speed of the skier when he leaves the end of the ramp is
2
2
v0 = v0 x + v0 y = ( +16.8 m/s ) + ( +14.1 m/s )
2 2 = 21.9 m/s The angle that the initial velocity makes with respect to the horizontal is v0 y −1 +14.1 m/s = tan = 40.0° v0 x +16.8 m/s ______________________________________________________________________________ θ = tan −1 156 KINEMATICS IN TWO DIMENSIONS 78. REASONING The relative velocities in this problem are:
vPS = velocity of the Passenger relative to the Shore
vP2 = velocity of the Passenger relative to Boat 2 (1.20 m/s, due east)
v2S = velocity of Boat 2 relative to the Shore
v21 = velocity of Boat 2 relative to Boat 1 (1.60 m/s, at 30.0° north of east)
v1S = velocity of Boat 1 relative to the Shore (3.00 m/s, due north)
The velocity vPS of the passenger relative to the shore is related to vP2 and v2S by (see the
method of subscripting discussed in Section 3.4):
vPS = vP2 + v2S
But v2S, the velocity of Boat 2 relative to the
shore, is related to v21 and v1S by
+y (North) v2S = v21 + v1S v1S
vPS 3.00 m/s Substituting this expression for v2S into the first
equation yields
vPS = vP2 + v21 + v1S
1.60 m/s This vector sum is shown in the diagram. We will
determine the magnitude of vPS from the equation
above by using the method of scalar components. 1.20 m/s 30.0° v21
+x (East) vP2 SOLUTION The table below lists the scalar components of the four vectors in the drawing.
Vector x Component y Component vP2 +1.20 m/s 0 m/s v21
v1S
vPS = vP2 + v21 + v1S +(1.60 m/s) cos 30.0° = +1.39 m/s +(1.60 m/s) sin 30.0° = +0.80 m/s
0 m/s +3.00 m/s +1.20 m/s + 1.39 m/s = +2.59 m/s +0.80 m/s +3.00 m/s = +3.80 m/s The magnitude of vPS can be found by applying the Pythagorean theorem to its x and y
components: vPS = ( 2.59 m/s ) + ( 3.80 m/s ) = 4.60 m/s
_______________________________...
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