7. For a certain interval of motion, the pin P is forced to move in the fixed parabolic

slot by the vertical slotted guide, which moves in the x direction at the constant rate

of 40 mm/s. All measurements are in mm and s. Calculate the magnitudes of Ԧ𝑣 and

Ԧ𝑎 of pin P when x = 60 mm.

SOLUTION

Trajectory of the pin is160

2xy ??60 avmmxwhen

jaiaajvivv yxyx

jaiaajvivv yxyx

yaxayvxv yxyx ,,

0)(/40 xacstsmmxv xx

The first derivative of trajectory equation with respect to time

smm

xxxxy

xy /30

80

4060

80160

2

160

2

jiv

3040 Magnitude of velocity of the pin: smmv /503040 22

22

2 /2080

40

80

1

80smmxxxy

xx

dt

dy

dt

d

ja

20

Magnitude of acceleration of the pin:

2/20 smma

8. Pins A and B must always remain in the vertical slot of yoke C, which

moves to the right at a constant speed of 6 cm/s. Furthermore, the pins

cannot leave the elliptic slot. What is the speed at which the pins approach

each other when the yoke slot is at x = 50 cm? What is the rate of change

of speed toward each other when the yoke slot is again at x = 50 cm?

100 cm

60 cm

x

6 cm/s

yoke

Cx

y

10. A long-range artillery rifle at A is aimed at an angle of 45o with the

horizontal, and its shell is just able to clear the mountain peak at the top of its

trajectory. Determine the magnitude u of the muzzle velocity, the height H of the

mountain above sea level, and the range R to the sea.

x

y

Horizontal:

Vertical:

SOLUTION

tvxxuvvaxooxoxx 45cos0

2

2

145sin gttvyygtugtvvga

yooyoyy

B

C

Point B (at apex) BByoy tutgugtvv 873.13045sin0

81.9

1

B

Bxoot

ututvxx11313.7

45cos08000

21 = smustB /18.39655.28

22 55.2881.92

155.2845cos18.3960

2

1 Byoo ygttvyy

myHmy BB 46006004000

Point C (sea level)

060014.280905.4

81.92

145cos18.3960600

2

1

2

2

2

CC

CC

yoo

tt

tt

gttvyy

06.2

18.59

Ct

mR

R

tvxxxoo

78.16578

18.5845cos18.3960

11. A projectile is launched with an initial speed of 200 m/s at an angle

of 60o with respect to the horizontal. Compute the range R as measured

up the incline.

x

y

s/m.sinv

s/mcosv

yo

xo

217360200

10060200

Horizontal:

tcosRtvxx xoo 100020 (1) at B

Vertical:

ay=gax=0

22 8192

12173020

2

1t.t.sinRgttvyy yoo (2) at B

(1) t=0.0094R

(2)

m.R

R..R..sinR

22967

009408192

100940217320

2

SOLUTION

12 . Determine the location h of the spot which the pitcher must throw

if the ball is to hit the catcher’s mitt. The ball is released with a speed of

40 m/s.

x

y

sinv

cosv

yo

xo

40

40

Horizontal:

cos

ttcostvxx xoo2

14020

Vertical:

ay=gax=0

22 8192

140081

2

1t.tsin.gttvyy yoo

586

6381

35516

02860

22612

574022614202005740202261

0812012261

1226120812

1819

2

1

2

14081

2

212

2

2222

.

.

.

.x

.

..x.xx.

xtan.tantan.

tansecsec.tan.cos

.cos

sin.

,

20 m

d1.638o

d=20 tan1.638=0.572 mh=2.8(1+0.572)=1.228 m

d

SOLUTION

13. A golfer strikes the ball at the top of a hill with an initial velocity of vo=12 m/s

and =60o with the horizontal. Determine the coordinates of the point the ball hits

the ground, its velocity at this moment and its total flight time. The equation of

curvature of the hill is given by y= 0.05x2. Dimensions are given in meters. Take

the gravitational attraction of the earth as g=9.81 m/s2.

vo

x

y=0.05x2

y

B

Determine the coordinates of the point the ball hits the ground, its velocity at this moment and its total

flight time.

vo=12 m/s

x60o

y=0.05x2

y

O

B

For B

For O

(t2 is the total flight time.)

Coordinates of B: xB=20.1 m , yB=20.24 m

Impact velocity: vx=6 m/s (cst), vy=(vo)y gt=10.39 9.81(3.35)=22.47 m/s

s/m.sinv

s/mcosv

yo

xo

39106012

66012

Horizontal: ttvxx xoo 6

Vertical:

ay=gax=0

2

2

90543910

2

1

t.t.y

gttvyy yoo

st

tttttt

tttxy

35.3

00105.339.1008.1905.439.10

605.0905.439.1005.0

2

1

222

222

s/m.vvv yx 262322