y
Problem 3.23
! = 30
A student throws a rock 30 degrees
below the horizontal moving with
velocity magnitude 18 m/s. The cliff is
50 meters high.
A student throws a rock 30 degrees below the horizontal
moving with velocity magnitude 18 m/s. The cliff is 50
meters high.
o
vo = 18 m/s
! = 30 o
vo = 18 m/s
c.) Write out the velocity as a function of time
for both the y and x direction.
a.) What are the stone’s initial coordinates?
x
d.) Write out the position as a function of time for both the y and x direction.
b.) What are the components of the initial velocity?
1.)
y
A student throws a rock 30 degrees below the horizontal
moving with velocity magnitude 18 m/s. The cliff is 50
meters high.
! = 30
3.)
A student throws a rock 30 degrees below the horizontal
moving with velocity magnitude 18 m/s. The cliff is 50
meters high.
o
vo = 18 m/s
vo = 18 m/s
c.) Write out the velocity as a function of time
for both the y and x direction.
a.) What are the stone’s initial coordinates?
(0, 50 m)
0
v2,x = vo,x + a x !t
b.) What are the components of the initial velocity?
x
! = 30
= (18 m/s ) cos 30 o
(
)
= ( "9 m/s ) + "9.8 m/s 2 !t
o
0
0
" 1%
2
x 2 = x o + vo,x !t + $ ' a x ( !t )
# 2&
vo,y
vo,y = !vo sin "
= 15.6 m/s
v2,y = vo,y + a y !t
d.) Write out the position as a function of time for both the y and x direction.
vo,x
vo,x = vo cos!
= 15.6 m/s
! = 30 o
vo = 18 m/s
= (15.6 m/s ) !t
= ! (18 m/s ) sin 30 o
" 1%
2
y 2 = y o + vo,y !t + $ ' a y ( !t )
# 2&
" 1%
2
= ( 50 m ) + ( (9 m/s ) !t + $ ' (9.8 m/s 2 ( !t )
# 2&
(
)
= !9 m/s
Note I had to manually put the negative sign in for the y-component of the velocity.
2.)
4.)
A student throws a rock 30 degrees below the horizontal
moving with velocity magnitude 18 m/s. The cliff is 50
meters high.
! = 30
A student throws a rock 30 degrees below the horizontal
moving with velocity magnitude 18 m/s. The cliff is 50
meters high.
o
vo = 18 m/s
e.) How long does it take for the stone to
reach the ground below?
! = 30 o
vo = 18 m/s
f. (con’t.) With what speed and angle of
impact does the stone land?
For the y-direction:
vgr,x
vgr,x
(v ) = (v )
2
gr,y
vgr,y
vgr
o,y
2
(
)
+ 2 ( !9.8 m/s ) ( 0 ! ( 50 m ))
+ 2a y y gr ! y o
= ( !9 m/s )
2
2
vgr,y
vgr
" vgr,y = !32.6 m/s
f.) With what speed and angle of impact does the stone land?
Note that the negative sign was inserted manually. Also, you could also have used:
vgr,y = v1,y + a y !t
(
)
= ( "9 m/s ) + "9.8 m/s 2 ( 2.4 s )
# vgr = "32.6 m/s
5.)
A student throws a rock 30 degrees below the horizontal
moving with velocity magnitude 18 m/s. The cliff is 50
meters high.
f. (con’t.) With what speed and angle of
impact does the stone land?
! = 30 o
vo = 18 m/s
e.) How long does it take for the stone to
reach the ground below?
((
vgr = vgr,x
vgr,x
" 1%
2
0 = ( 50 m ) + ( )9 m/s ) !t + $ ' )9.8 m/s 2 ( !t )
# 2&
(
! = 30 o
vo = 18 m/s
To get the net velocity, the Pythagorean relationship
holds:
" 1%
2
y 2 = y o + vo,y !t + $ ' a y ( !t )
# 2&
(
7.)
)
vgr
( t = 2.4 seconds
(
) + (v ) )
2 1/2
2
gr,y
= (15.6 m/s ) + ( !32.6 m/s )
" vgr = 36.1 m/s
vgr,y
at an angle of:
f.) With what speed and angle of impact does the stone land?
For the x-direction:
vgr,x
)
2 1/2
vgr
vgr,y
" vgr,y %
! = tan $
'
# vgr,x &
" (32.6 m/s %
= tan $
# 15.6 m/s '&
0
v2,x = vo,x + a x ( !t )
= (15.6 m/s )
This shouldn’t be surprising as the velocity in the x-direction DOESN’T CHANGE unless
there’s a force in the x-direction, which there isn’t in this case!
2
so:
) ! = (64.4 o
!
vgr = ( 36.1 m/s ) " ! 64.4 o
6.)
8.)