Download PROBLEM 11.99

PROBLEM 11.99
A baseball pitching machine
“throws”
baseballs
with
a
horizontal velocity v0. Knowing
that height h varies between 788
mm and 1068 mm, determine
(a) the range of values of v0, (b) the
values of
corresponding to
h 788 mm and h 1068 mm.
SOLUTION
(a)
y0
Vertical motion:
y
y0
(v y )0 t
1 2
gt
2
or
y
At Point B,
0
2( y0 y)
g
t
or
h
1.5 m, (v y )0
tB
2( y0 h)
g
When h
788 mm
0.788 m,
tB
(2)(1.5 0.788)
9.81
0.3810 s
When h
1068 mm
1.068 m,
tB
(2)(1.5 1.068)
9.81
0.2968 s
Horizontal motion:
x0
x
0, (vx )0
v0t
v0 ,
or
v0
x
t
xB
tB
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.99 (Continued)
12.2 m,
With xB
we get
and
32.02 m/s
(b)
v0
41.11 m/s
v0
12.2
0.3810
32.02 m/s
v0
12.2
0.2968
41.11 m/s
115.3 km/h
or
Vertical motion:
vy
(v y )0
Horizontal motion:
vx
v0
tan
dy
dx
(v y ) B
(vx ) B
gt
v0
148.0 km/h
gt
gt B
v0
For h
0.788 m,
tan
(9.81)(0.3810)
32.02
0.11673,
6.66
For h
1.068 m,
tan
(9.81)(0.2968)
41.11
0.07082,
4.05
PROBLEM 11.100
While delivering newspapers, a
girl throws a newspaper with a
horizontal velocity v0. Determine
the range of values of v0 if the
newspaper is to land between
Points B and C.
SOLUTION
Vertical motion. (Uniformly accelerated motion)
y
0 (0)t
1 2
gt
2
Horizontal motion. (Uniform)
x 0 (vx )0 t
At B:
y:
tB
or
Then
x:
y:
or
2 ft
tC
or
Then
7 ft
(v0 ) B
or
At C:
1
3 ft
3
x:
1
12 ft
3
(v0 )C
v0t
1
(32.2 ft/s 2 )t 2
2
0.455016 s
(v0 ) B (0.455016 s)
15.38 ft/s
1
(32.2 ft/s 2 )t 2
2
0.352454 s
(v0 )C (0.352454 s)
35.0 ft/s
15.38 ft/s v0
35.0 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.103
A volleyball player serves the
ball with an initial velocity v0 of
magnitude 13.40 m/s at an angle
of 20° with the horizontal.
Determine (a) if the ball will
clear the top of the net, (b) how
far from the net the ball will land.
SOLUTION
First note
(a)
(v x ) 0
(13.40 m/s) cos 20
12.5919 m/s
(v y ) 0
(13.40 m/s) sin 20
4.5831 m/s
Horizontal motion. (Uniform)
0 (vx ) 0 t
x
9 m (12.5919 m/s) t or tC
At C
0.71475 s
Vertical motion. (Uniformly accelerated motion)
y
At C:
yC
y0
(v y ) 0 t
1 2
gt
2
2.1 m (4.5831 m/s)(0.71475 s)
1
(9.81 m/s 2 )(0.71475 s) 2
2
2.87 m
2.43 m (height of net)
yC
(b)
At B, y
0:
0
ball clears net
1
(9.81 m/s 2 )t 2
2
2.1 m (4.5831 m/s)t
Solving
tB
1.271175 s (the other root is negative)
Then
d
(v x ) 0 t B
(12.5919 m/s)(1.271175 s)
16.01 m
The ball lands
b
(16.01 9.00) m
7.01 m from the net