Quiz1 - Phys
2101
Sec6on
7
...

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Unformatted text preview: Phys
2101
Sec6on
7
 Quiz
1
–
Form
A
 
Aug
31
2010
 A
plane,
diving
with
constant
speed
at
an
angle
of
60°
with
the
ver8cal,
releases
a
 projec8le
at
an
al8tude
of
700
m.
The
projec8le
hits
the
ground
6
s
a?er
release.
 (a)  What
is
the
speed
of
the
plane?

 In
the
drawing
above,
the
angle
of
the
ini8al
velocity
vector
with
the
posi8ve
 horizontal
x‐axis
is
θ=-30o,
and
the
ini8al
height
is
y0=700m.

We
know
that
at
 t=6s,
the
projec8le
is
on
the
gorund,
which
we
will
take
as
y=0.

 The
equa8on
for
the
y‐component
of
the
posi8on
vector
is

 1 y = y 0 + v 0 sin θ t − gt 2 2 so
at
t=6s,
when
y=0,
we
know
everything
but
the
ini8al
speed:


 1 0 = y 0 + v 0 sin θ t − gt 2 ⇒ € 2 12 gt − y 0 0.5 × 9.8 m / s2 × (6 s) 2 − 700 m −523.6 m 2 v0 = = = = 174.53 m / s sin θ t sin(−30 o ) × 6 s −3s € (No8ce
that
we
cannot
get
a
nega8ve
number
for
the
ini8al
speed
unless
we
do
 something
wrong…)
 (b)
What
is
the
projec6le’s
speed
when
it
hits
the
ground?

 We
need
x
and
y
components
of
the
velocity
at
t=6s:
 v x = v 0 cosθ = 174.53 m/s × cos(−30 o ) = 151.15 m/s v y = v 0 sin θ − gt = 174.53 m/s × sin(−30 o ) − 9.8m/s2 × 6 s = −146.07 m/s v = v x 2 + v y 2 = 210.19 m/s € Phys
2101
Sec6on
7
 Quiz
1
–
Form
B
 
Aug
31
2010
 A
plane,
diving
with
constant
speed
at
an
angle
of
50°
with
the
ver8cal,
releases
a
 projec8le
at
an
al8tude
of
500
m.
The
projec8le
hits
the
ground
5
s
a?er
release.
 (a)
What
is
the
speed
of
the
plane?

 In
the
drawing
above,
the
angle
of
the
ini8al
velocity
vector
with
the
posi8ve
 horizontal
x‐axis
is
θ=-40o,
and
the
ini8al
height
is
y0=500m.

We
know
that
at
 t=5s,
the
projec8le
is
on
the
ground,
which
we
will
take
as
y=0.

 The
equa8on
for
the
y‐component
of
the
posi8on
vector
is

 1 y = y 0 + v 0 sin θ t − gt 2 2 so
at
t=5s,
when
y=0,
we
know
everything
but
the
ini8al
speed:

 € 1 0 = y 0 + v 0 sin θ t − gt 2 ⇒ 2 12 gt − y 0 0.5 × 9.8 m / s2 × (5 s) 2 − 500 m −377.5 m v0 = 2 = = = 117.6 m / s sin θ t sin(−40 o ) × 5 s −3.21s (b)
How
far
does
the
projec8le
travel
horizontally
during
its
flight?
 € Since
we
know
the
projec8le
travels
with
constant
velocity
in
the
horizontal
 direc8on,
we
know
that
distance
=
velocity
*
8me:

 d = v 0 x t = v 0 cos θ t = 117.6 m / s × cos(−40 o ) × 5 s = 450 m € ...
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