Department of Natural Sciences
Clayton College & State University
September 16, 2002
Physics 1111 – Quiz 3
Name _____________________________________
1. You throw a physics textbook horizontally at a speed of 9.00 m/s from a top of a
building. The height of the building is 30.0 m. Ignoring air resistance,
a. What are the components of the initial velocity of the book?
V0: 9.00 m/s @ 0.0o
V0x = V0 cos(0) = 9.00m/s cos(0o) = 9.00 m/s
V0y = V0 sin(0) = 9.00m/s sin(0o) = 0 m/s
b. How long will it take for the book to hit the ground?
y0 = 30.0 m
y=0m
y = y0 + V0y t – ½ g t 2
0 m = 30.0 m + (0 m/s) t – ½ (9.80 m/s2)t 2
½ (9.80 m/s2)t 2 = 30.0 m
t2 = 30.0 m/(½ (9.80 m/s2))
t = 2.47 s
c. How far from the side of the building do you need to walk in horizontal
direction to retrieve the textbook?
x0 = 0 m
x = x0 + V0x t
x = 0 m + (9.00 m/s)(2.47 s)
x = 22.3 m
Clayton State University
Department of Natural Sciences
September 27, 2005
Physics 1111 – Quiz 5.1
Name ___SOLUTION__________________________________
1. You throw a physics textbook with a speed of 9.00 m/s at 25.0o angle from a top
of a building. The textbook lands on the ground 2.50 s later. Ignoring air
resistance,
a. What are the components of the initial velocity of the book?
V0: 9.00 m/s @ 25.0o
Vx0 = V0 cos(0) = (9.00 m/s) cos(25.0o) = 8.16 m/s
Vy0 = V0 sin(0) = (9.00 m/s) sin(25.0o) = 3.80 m/s
b. How high is the building?
y = y0 + Vy0 t – ½ g t2
0 = y0 + Vy0 t – ½ g t2
y0 = - Vy0 t + ½ g t2
y0 = 21.2 m
c. How far from the side of the building do you need to walk in horizontal
direction to retrieve the textbook?
x = x0 + Vx0 t
x = (0 m) + (8.16 m/s) (2.50 s) =20.4 m
Clayton State University
Department of Natural Sciences
September 25, 2006
Physics 1111 – Quiz 4
Name __SOLUTION___________________________________
1. A soccer ball is kicked off the 40.0-m cliff at a speed of 20.0 m/s horizontally.
a. What are the components of the initial velocity of the ball?
V0: 20.0 m/s @ 0o
Vx0 = V0 cos(0) = (20.0 m/s) cos(0o) = 20.0 m/s
Vy0 = V0 sin(0) = (20.0 m/s) sin(0o) = 0 m/s
b. How long is the ball in the air?
y = y0 + Vy0 t – ½ g t2
0 = y0 + 0 – ½ g t2
½ g t2 = y0
g t2 = 2 y0
t2 = 2 y0 /g
t = (2 y0 /g)1/2
t = (2 (40.0 m) /(9.81 m/s2))1/2
t = 2.86 s
c. How far from the base of the cliff do you need to walk in horizontal
direction to retrieve the ball?
x = x0 + Vx0 t
x = (0 m) + (20.0 m/s) (2.86 s) =57.2 m
Department of Natural Sciences
Clayton College & State University
February 9, 2004
Physics 1111 – Quiz 4
Name _____________________________________
1. A crow is flying horizontally with a constant speed of 2.70 m/s when it releases a
clam from its beak. The clam lands on the rocky beach 2.10 s later. Just before the
clam lands, what is
a. Its horizontal component of velocity?
Initial velocity: V0: 2.70 m/s @ 0o
V0x = V0 cos(0) = (2.70 m/s) cos(0o) = 2.70 m/s
V0y = V0 sin(0) = (2.70 m/s) sin(0o) = 0 m/s
Vx = V0x = 2.70 m/s
b. Its vertical component of velocity?
Vy = V0y - g t = 0 m/s – (9.80 m/s2) (2.10 s) = -20.6 m/s
c. From what height was the clam released?
y=0m
y = y0 + V0yt – ½ g t2
0 = y0 – ½ g t2
y0 = ½ g t2
y0 = 21.6 m
d. What is the range of the clam’s trajectory?
x0 = 0
x = x0 + V0x t
x = 0 m + (2.70 m/s) (2.10 s)
x = 5.67 m
Department of Natural Sciences
Clayton College & State University
February 9, 2004
Physics 1111 – Quiz 4
Name ______SOLUTION_______________________________
A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s
and accidentally drops a bomb (fortunately not armed) at an elevation of 300 m. You can
ignore air resistance.
a. What are the components of the initial velocity?
V0: 60.0 m/s @ 0o
Vx0 = V0 cos(0) = (60.0 m/s) cos(0o) = 60.0 m/s
Vy0 = V0 sin(0) = (60.0 m/s) sin(0o) = 0 m/s
b. How much time is required for the bomb to reach the ground?
y0 = 300 m
y=0
y = y0 + Vyo t – ½ g t2
0 = y0 – ½ g t2
y0 = ½ g t2
t2 = 2 y0 /g
t = (2 y0 /g)1/2
t = 7.82 s
c. How far does it travel in horizontal direction while falling?
x = x0 + Vx0 t
x = (0 m) + (60.0 m/s) (7.82 s) = 469 m
Department of Natural Sciences
Clayton State University
February 7, 2007
Physics 1111 – Quiz 4
Name __SOLUTION___________________________________
1. You throw a physics textbook horizontally at a speed of 11.0 m/s from a top of a
building. The height of the building is 50.0 m. Ignoring air resistance,
a. What are the components of the initial velocity of the book?
V0: 11.0 m/s @ 0o
Vx0 = V0 cos(0) = (11.0 m/s) cos(0o) = 11.0 m/s
Vy0 = V0 sin(0) = (11.0 m/s) sin(0o) = 0 m/s
b. How long will it take for the book to hit the ground?
y = y0 + Vy0 t – ½ g t2
0 = y0 + 0 – ½ g t2
½ g t2 = y0
g t2 = 2 y0
t2 = 2 y0 /g
t = (2 y0 /g)1/2
t = (2 (50.0 m) /(9.81 m/s2))1/2
t = 3.19 s
c. How far from the side of the building do you need to walk in horizontal
direction to retrieve the textbook?
x = x0 + Vx0 t
x = (0 m) + (11.0 m/s) (3.19 s) =35.1 m
Department of Natural Sciences
Clayton State University
February 13, 2008
Physics 1111 – Quiz 5
Name _____________________________________
A projectile is fired with an initial speed of 35.0 m/s at an angle of 45.0o above the
horizontal on a long flat firing range. Determine
a. The components of the projectile’s initial velocity.
Vx0 = V0 cos(0) = (35.0 m/s) cos(45.0o) = 24.7 m/s
Vy0 = V0 sin(0) = (35.0 m/s) sin(45.0o) = 24.7 m/s
b. The time of flight.
y = y0 + Vy0 t – ½ gt2
0 = 0 + Vy0 t – ½ gt2
0 = t (Vy0 – ½ gt )
Vy0 – ½ gt = 0
t = 2 Vy0 /g
t = 5.05 s
c. The projectile’s range.
x = x0 + Vx0 t
x = (0 m) + (24.7 m/s) (5.05 s) = 125 m
Department of Natural Sciences
Clayton College & State University
June 8, 2004
Physics 1111 – Quiz 5
Name ____SOLUTION_________________________________
A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an
angle of 36.9o above the horizontal. You can ignore air resistance.
a. What are the components of the initial velocity?
V0: 30.0 m/s @ 36.9o
Vx0 = V0 cos(0) = 30.0m/s cos(36.9.0o) = 24.0 m/s
Vy0 = V0 sin(0) = 30.0m/s sin(36.9o) = 18.0 m/s
b. The ball is caught at the same elevation as it was hit. How much time was the ball
in the air?
y = y0 + V0yt – ½ g t2
0 = 0 + V0yt – ½ g t2
0 = t (V0y – ½ g t)
(V0y – ½ g t) = 0
2V0y /g = t
t = 3.68 s
c. How far did the ball travel in horizontal direction?
x = x0 + Vx0 * t
x = 0 + (24.0 m/s) * (3.68 s) = 88.2 m
d. What is the velocity of the ball at the top of its trajectory?
Vx = Vx0 = 24.0 m/s
Vy = 0 m/s
V: 24.0 m/s @ 0o
Department of Natural Sciences
Clayton State University
June 15, 2005
Physics 1111 – Quiz 4
Name ___SOLUTION__________________________________
A famous soccer player, Hedley Footy, kicks the ball at a speed of 15.0 m/s when t = 0 s.
The initial velocity vector of the ball makes an angle of 60.0o with the horizontal
direction.
a. Find the components of the initial velocity.
V0: 15.0 m/s @ 60.0o
Vx0 = V0 cos(0) = (15.0 m/s) cos(60.0o) = 7.50 m/s
Vy0 = V0 sin(0) = (15.0 m/s) sin(60.0o) = 13.0 m/s
b. If the ball lends at the same level as it started, how long is the ball in
motion?
y = y0 + Vy0 t - 1/2 g t2
y = y0 = 0
0 = Vy0 t - 1/2 g t2 = t ( Vy0 - 1/2 g t)
Vy0 - 1/2 g t = 0
2 Vy0 / g = t
t = 2 (13.0 m/s) / (9.80 m/s2) = 2.65 s
c. What is the projectile’s range?
x = x0 + Vx0 t
x = (0 m) + (7.50 m/s)(2.65 s) = 19.9 m
d. What is the ball’s speed at the top of trajectory?
V = 7.50 m/s
Remember, Vy0 = 0 at the top!
Department of Natural Sciences
Clayton State University
June 18, 2007
Physics 1111 – Quiz 5
Name ___SOLUTION__________________________________
A soccer ball is kicked with a speed of 9.50 m/s at an angle of 40.0o above the horizontal
direction. Some time later the ball lands at the same level from which it was kicked.
a. Find the components of the ball’s initial velocity.
Vx0 = V0 cos(0) = (9.50m/s) cos(40.0o) = 7.28 m/s
Vy0 = V0 sin(0) = (9.50m/s) sin(40.0o) = 6.11 m/s
b. What are the components of the ball’s velocity at the top of the
trajectory?
Vx = Vx0 = 7.28 m/s
Vy = 0 m/s
c. How long does it take the ball to reach the top of the trajectory?
Vy = Vy0 – gt
(Vy0 - Vy) /g = t
t = (6.11 m/s – 0 m/s)/(9.80 m/s2) = 0.623 s
d. How far does the ball go in the horizontal direction?
x = x0 + Vx0 (t) = 0 m + (7.28 m/s) (1.25 s) = 9.07 m