Physics 40A  Midterm Exam Solutions  Fall 2005  Exam A
Problem 1
[25 points]
A stone is projected at a cliff of height
h
with an initial speed of
42
m/s at an angle of 60° above
the horizontal, as shown below. The stone strikes at point
A
, 5.5 s after it was launched. Find (a)
the height
h
of the cliff, (b) the speed of the stone just before impact at
A
, and (c) the maximum
height
H
reached above the ground.
Solution:
This is a projectile motion problem so the horizontal acceleration is zero and the vertical
acceleration is
−
g
. We define the origin of our coordinate system at the release point with
+
y
vertically upward and +
x
horizontal (toward the cliff). The
x
and
y
components of the initial
velocity are
v
0
x
=
v
0
cos
0
=
42 m/s
cos60
°
=
21.0 m/s
v
0
y
=
v
0
sin
0
42 m/s
sin60
°
36.37 m/s
(a) To find the height
h
we can use
y
y
0
=
v
0
y
t
1
2
gt
2
with
y
y
0
=
h
and
t
= 5.5 s. This gives
h
36.37 m/s
5.5 s
1
2
9.8 m/s
2
5.5 s
2
h
=
51.8 m
(b) The horizontal motion is motion with constant velocity, so
v
x
=
v
0
x
=
21.0 m/s
2
The vertical component of the velocity at impact is given by
v
y
=
v
0
y
=
36.37 m/s
9.8 m/s
2
5.5 s
=
17.53 m/s
Therefore, the speed at impact is
s
=
v
x
2
v
y
2
=
21.0 m/s
2
17.53 m/s
2
s
=
27.4 m/s
(c) We can use
v
y
2
=
v
0
y
2
2
g
y
y
0
At the maximum height
v
y
=
0
and we have
y
y
0
=
H
. Therefore, solving for
H
we
get
H
=
v
0
y
2
2
g
=
36.37 m/s
2
2
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 Winter '08
 Ellison
 Physics, Force, Friction, m/s, Midterm Exam Solutions

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