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Physics 40A  Midterm Exam Solutions  Fall 2008  Exam A
Problem 1
[25 points]
You throw a ball with a speed of
25
m/s at an angle of 40° above the horizontal directly toward a
wall (see Figure). The wall is 22 m from the release point of the ball. (a) How long does the ball
take to reach the wall? (b) How far above the release point does the ball hit the wall? (c) What are
the horizontal and vertical components of its velocity as it hits the wall? (d) When it hits, has it
passed the highest point on its trajectory? Explain your reasoning.
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 wall). The
x
and
y
components of the initial
velocity are
v
0x
=
v
0
cos
0
=
25 m/s
cos 40
°
=
19.2 m/s
v
0y
=
v
0
sin
0
25 m/s
sin 40
°
16.1 m/s
(a) To find the time at which the projectile hits the wall (
x
= 22 m) we can use
x
−
x
0
=
v
0x
t
The horizontal displacement of the ball is
x
−
x
0
= 22 m, so we obtain
t
=
x
−
x
0
v
0x
=
22 m
19.2 m/s
t
=
1.15s
(b) The vertical distance is
y
−
y
0
=
v
0y
t
−
1
2
gt
2
16.1 m/s
1.15 s
−
1
2
9.8 m/s
2
1.15 s
2
y
−
y
0
=
12.0 m
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This note was uploaded on 04/16/2008 for the course PHYS 40 taught by Professor Ellison during the Winter '08 term at UC Riverside.
 Winter '08
 Ellison
 Physics

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