Physics 107 HOMEWORK ASSIGNMENT #2

Physics 107 HOMEWORK ASSIGNMENT #2 Cutnell & Johnson, 7th edition Chapter 1: Problem 50 Chapter 2: Problems 44, 54, 56 Chapter 3: Problem 38 *50 Multiple-Concept Example 9 deals with the concepts that are important in this problem. A grasshopper makes four jumps. The displacement vectors are (1) 27.0 cm, due west; (2) 23.0 cm, 35.0° south of west; (3) 28.0 cm, 55.0° south of east; and (4) 35.0 cm, 63.0° north of east. Find the magnitude and direction of the resultant displacement. Express the direction with respect to due west. 44 Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver springs upward with an initial speed of 1.8 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = −3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water? *54 A ball is thrown upward from the top of a 25.0-m-tall building. The ball’s initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 31.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building? **56 While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.20 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction. *38 Multiple-Concept Example 4 deals with a situation similar to that presented here. A marble is thrown horizontally with a speed of 15 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 65° with the horizontal. From what height above the ground was the marble thrown?

SOLUTIONS: 50. REASONING The following table shows the components of the individual displacements and the components of the resultant. The directions due east and due north are taken as the positive directions. East/West Component

Displacement (1) (2) (3) (4)

North/South Component

–27.0 cm 0 –(23.0 cm) cos 35.0° = –18.84 cm –(23.0 cm) sin 35.0° = –13.19 cm (28.0 cm) cos 55.0° = 16.06 cm –(28.0 cm) sin 55.0° = –22.94 cm (35.0 cm) cos 63.0° = 15.89 cm (35.0 cm) sin 63.0° = 31.19 cm

Resultant

–13.89 cm

–4.94 cm

SOLUTION a. From the Pythagorean theorem, we find that the magnitude of the resultant displacement vector is 13.89 cm

R = (13.89 cm) 2 + (4.94 cm) 2 = 14.7 cm

R

b. The angle θ is given by

θ = tan −1

FG 4.94 cm IJ = H 13.89 cm K

θ

4.94 cm

19 .6 ° , south of west

44. REASONING AND SOLUTION a. v=±

v 2 = v02 + 2ay

(1.8 m/s )2 + 2 ( –9.80 m/s 2 ) ( –3.0 m ) = ±7.9 m/s

The minus is chosen, since the diver is now moving down. Hence, v = −7.9 m/s . b. The diver's velocity is zero at his highest point. The position of the diver relative to the board is y=–

v02 2a

=–

(1.8 m/s )2

(

2 –9.80 m/s 2

)

= 0.17 m

The position above the water is 3.0 m + 0.17 m = 3.2 m .

54. REASONING AND SOLUTION building at any time t is

The position of the ball relative to the top of the

y = v0t + 12 a t 2 = v0t +

1 2

( −9.80 m/s2 ) t 2

For the particular time that the ball arrives at the bottom of the building, y = –25.0 m. The equation becomes 4.90t2 – 12.0t – 25.0 = 0 The quadratic equation yields solutions t = 3.79 s and –1.34 s. The negative solution is rejected as being non-physical. During time t the person has run a distance x = vt so

v=

56. REASONING

x 31.0 m = = 8.18 m/s t 3.79 s

To find the initial velocity v0,2 of the second stone, we will employ

Equation 2.8, y = v0,2t2 +

1 at 2 . 2 2

In this expression t2 is the time that the second stone is in

the air, and it is equal to the time t1 that the first stone is in the air minus the time t3.20 it takes for the first stone to fall 3.20 m: t2 = t1 − t3.20

We can find t1 and t3.20 by applying Equation 2.8 to the first stone.

SOLUTION To find the initial velocity v0,2 of the second stone, we employ Equation 2.8, y = v0,2t2 + 12 at2 2 . Solving this equation for v0,2 yields

v0,2 =

y − 12 at2 2 t2

The time t1 for the first stone to strike the ground can be obtained from Equation 2.8,

y = v0,1t1 +

1 at 2 . 2 1

Noting that v0,1 = 0 m/s since the stone is dropped from rest and solving

this equation for t1, we have

t1 =

2y = a

2 ( −15.0 m ) = 1.75 s −9.80 m/s 2

(1)

Note that the stone is falling down, so its displacement is negative (y = − 15.0 m). Also, its 2 acceleration a is that due to gravity, so a = −9.80 m/s . The time t3.20 for the first stone to fall 3.20 m can also be obtained from Equation 1: t3.20 =

2y 2 ( −3.20 m ) = = 0.808 s a −9.80 m/s 2

The time t2 that the second stone is in the air is t2 = t1 − t3.20 = 1.75 s − 0.808 s = 0.94 s The initial velocity of the second stone is

v0,2 =

y − 12 at2 2

( −15.0 m ) − 12 ( −9.80 m/s 2 ) ( 0.94 s )2

= = −11 m/s t2 0.94 s ____________________________________________________________________

38. REASONING Since the vertical height is asked for, we will begin with the vertical part of the motion, treating it separately from the horizontal part. The directions upward and to the right are chosen as the positive directions in the drawing. The data for the vertical motion are summarized in the following table. Note that the initial velocity component v0y is zero, because the marble is thrown horizontally. The vertical component y of the marble’s displacement is entered in the table as −H, where H is the height we seek. The minus sign is included, because the marble moves downward in the negative y direction. The vertical component vy of the final velocity is checked as an important variable in the table, because we are given the angle that the final velocity makes

v0x vx 65° vy

H

vx

vy

y-Direction Data y

ay

vy

v0y

−H

−9.80 m/s2

√

0 m/s

t

with respect to the horizontal. Ignoring air resistance, we apply the equations of kinematics. With the data indicated in the table, Equation 3.6b becomes v 2y 2 2 2 v y = v0 y + 2a y y = ( 0 m/s ) + 2a y ( − H ) or H = − (1) 2a y

SOLUTION To use Equation (1), we need to determine the vertical component vy of the final velocity. We are given that the final velocity makes an angle of 65° with respect to the horizontal, as the inset in the drawing shows. Thus, from trigonometry, it follows that

tan 65° =

−v y vx

or

v y = −vx tan 65°

or

v y = −v0 x tan 65°

where the minus sign is included, because vy points downward in the negative y direction. In the absence of air resistance, there is no acceleration in the x direction, and the horizontal component vx of the final velocity is equal to the initial value v0x. Substituting this result into Equation (1) gives

H =−

v 2y 2a y

( −v0 x tan 65°) =− 2a y

2

2

 − (15 m/s ) tan 65°  =− = 53 m 2 −9.80 m/s 2

(

)