Projectile_Practice Key

Projectile Motion Practice
Answer Key
Directions: Show all of your work as though this were a test. That will make it easier to
find any mistakes you make and easier to explain how you did it to your confused friend.
1. Fred throws a ball straight up into the air. It rises for 1.5 seconds, and then falls back into his
hand 3.0 seconds after he first released it.
A) Make a graph of the position of the ball as a function of time.
Height (m)
Time (s)
B) Make a graph of the velocity of the ball as a function of time.
Velocity (m/s)
Time (s)
page 1 of 7
C) Make a graph of the acceleration of the ball as a function of time.
a (m/s2)
Time (s)
D) What is the initial velocity of the ball?
At the top of the ball’s flight, v = 0 m/s and t = 1.5 s.
v(t) = vo + at; a = –g
v = vo – gt
vo = v + gt
vo = 0 + (9.81 m/s2)(1.5s) = 14.715 m/s
vo = 15 m/s
E) What is the location of the ball at 2.0 s?
Find: y(2)
Given: yo = 0; vo = 15 m/s; a = –g
y(t) = yo + vot + ½at2
y(2) = 0 m + (14.715 m/s)(2.0 s) – ½g(2.0)2
y(2) = 0 m + 29.43 m – 19.62 m = 9.81 m/s
y(2) = 9.8 m
F) What is the velocity of the ball at 2.0 s?
Find: v(2)
v(t) = vo + at
v(2) = (14.715 m/s) – g(2.0 s)
v(2) = 14.715 m/s – 19.62 m/s = –4.905 m/s
v(2) = –4.9 m/s
G) What is the acceleration of the ball at 2.0 s?
a = –g = –9.81 m/s2 for all times.
page 2 of 7
2. Fred now places the ball on the ground and kicks it upward at an angle of 30° above the
level field with an initial speed of 36 m/s.
A) What is the velocity of the ball 2.5 s after it is kicked?
x
y
vx(t) = vox + axt
vy(t) = voy + ayt
vx(t) = vox + 0
vy(t) = voy – gyt
vx(2.5) = vocos(30°)
vy(2.5) = vosin(30°) – g(2.5 s)
vx(2.5) = (36 m/s)cos(30°)
vy(2.5) = (36m/s)sin(30°) – (9.81 m/s2)(2.5 s)
vx(2.5) = 31.17691…m/s
vy(2.5) = –6.525 m/s
v = (vx2 + vy2)½
v = 32 m/s
θ = arctan(vy/vx)
θ = –12° = 348°
B) Where is the ball 2.5 s after it is kicked?
x
y
x(t) = xo + voxt + ½axt2
y(t) = yo + voyt + ½ayt2
x(t) = 0 + vocos(30°)(2.5 s) + 0
y(t) = 0 + vosin(30°)(2.5 s) – ½g(2.5 s)2
x(t) = (36 m/s)cos(30°)(2.5 s)
y(t) = (36m/s)sin(30°)(2.5 s) – ½ (9.81 m/s2)(2.5 s)2
x(2.5) = 77.94228634 m
y(2.5) = 14.34375 m
The ball is about 78 m away from its starting point at a height of 14 m, or
r = (x2 + y2)½
θ = arctan(y/x)
r = 79 m
θ = 10°
The ball is at a displacement of 79 m at an angle of 10° above the horizontal.
C) What is the maximum height attained by the ball?
When the ball is at its maximum height, the y-component of its velocity is zero.
vy2 = voy2 + 2ay∆y
0 = voy2 – 2g∆y
2g∆y = voy2
∆y = voy2/2g
(y – yo) = voy2/2g
(y – 0) = (36sin(30°))2/2g = 16.5137…m
y = 17 m
page 3 of 7
D) How much time does the ball spend in the air before it first lands?
When the ball first bounces, y = yo = 0 m.
y(t) = yo + voyt + ½ayt2
0 = 0 + voyt – ½gt2
½gt2 = voyt
t = 2voy/g
t = 2(36 m/s)(sin(30°))/(9.81 m/s2) = 3.669724771 s
t = 3.7 s
E) How far does the ball travel across the field before it lands?
x(t) = xo + voxt + ½axt2
x(3.7) = 0 + (36 m/s)cos(30°)(3.669724771 s) + 0 = 114.4106956 m
x(3.7) = 110 m
F) What is the velocity of the ball as it first lands?
vf = 36 m/s at –30° = 330°; parabolas are symmetric.
3. Maggie, angry at her insensitive boyfriend, throws his picture in its frame off of a cliff. The
picture leaves her hand at an angle of +25° above the horizontal with a speed of 14.0 m/s.
The picture frame hits the rocks below 7.00 seconds after it leaves Maggie’s hand. Neglect
the affects of air resistance.
A) Where is the picture frame 2.00 seconds after it leaves Maggie’s hand?
to = 0 s
xo = 0 m
yo = 0 m
vo = 14 m/s at 25°
t = 7.00 s
xf = ? m
yf = ? m
vf = ??
page 4 of 7
x
y
x(t) = xo + voxt + ½axt2
y(t) = yo + voyt + ½ayt2
x(2) = 0 + (14.0 m/s)cos(25°)(2 s) + 0
y(2) = 0 + (14.0 m/s)sin(25°)(2 s) – ½g(2 s)2
x(2) = 25.37661804 m
y(2) = –7.786688671m
The frame is about 25.4 m away from its starting point at a height of 7.79 m below its
starting point, or
r = (x2 + y2)½
θ = arctan(y/x)
r = 26.5 m
θ = –17.1° = 343°
The ball is at a displacement of 26.5 m at an angle of 17.1° below the horizontal.
B) What is the velocity of the picture frame 2.00 seconds after it leaves Maggie’s hand?
x
y
vx(t) = vox + axt
vy(t) = voy + ayt
vx(t) = vox + 0
vy(t) = voy – gyt
vx(2) = (14.0 m/s)cos(25°)
vy(2) = (14.0 m/s)sin(25°) – g(2 s)
vx(2) = 12.68830902 m/s
vy(2) = –13.70334434 m/s
v = (vx2 + vy2)½
v = 18.7 m/s
θ = arctan(vy/vx)
θ = –47.2° = 313°
C) How high is the cliff?
y(t) = yo + voyt + ½ayt2
y(7) = 0 + (14.0 m/s)sin(25°)(7.00 s) + ½(9.81 m/s 2)(7.00 s)2
y(7) = 11.8 m – 240 m
y(7) = 229 m
D) How high above the cliff does the picture frame rise before starting its journey
downward to the rocks below?
Well, at the top of its trajectory the y-component of its velocity will be zero.
vy2 = voy2 + 2ay∆y
0 = voy2 – 2g∆y
2g∆y = voy2
∆y = voy2/2g
(y – yo) = voy2/2g
(y – 0) = (14.0 m/s)sin(25°))2/2g = 1.7841297… m
y = 1.78 m above the cliff.
page 5 of 7
E) What is the velocity of the picture frame as it reaches the rocks below?
x
y
vx(t) = vox + axt
vy2 = voy2 + 2ay∆y
vx(t) = vox + 0
vy2 = voy2 – 2g∆y
vx(7) = (14.0 m/s)cos(25°)
vy2 = (14.0 m/s)sin(25°))2 – 2g(–229 m – 0 m)
vx(7) = (14.0 m/s)cos(25°)
vy2 = (14.0 m/s)sin(25°))2 – 2g(–229 m – 0 m)
vy2 = 4527.986814 m 2/s 2
vx(7) = 12.7 m/s
v = (vx2 + vy2)½
v = 68.5 m/s
vy –67.3 m/s (negative root; it’s falling!)
θ = arctan(vy/vx)
θ = –79.3° = 281°
F) What is the horizontal distance traveled by the picture frame before it is smashed on the
rocks below?
x(t) = xo + voxt + ½axt2
x(7) = 0 + (14.0 m/s)cos(25°)(7.00 s) + 0 = 88.81816313 m
x(7) = 88.8 m
4. Lucy and Ethel team up to build a water balloon launcher. To test it, the set it up on top of
their apartment building, 50.0 meters above street level.
A) What angle of launch will give them maximum range?
The maximum range occurs at θo = 45°
B) If launched at the best angle with a speed of 20.0 m/s, where will the balloon land?
y(x) = (tan θo)x – [(g/2cos2 θo)/vo2]x2
–50.0 m = (tan 45°)x – [(g/2cos2 45°)/(20.0 m/s)2]x2
–50.0 m = (1)x – [(g/2cos2 45°)/(20.0 m/s)2]x2
–50.0 m = x – [(9.81 m/s2)/(400 m2/s2)]x2
–50.0 m = x – [0.024525 m]x2
[0.024525 m]x2 – x – 50.0 m = 0
x = –29.2 m , +69.9 m
x = +69.9 m Choose the positive root; the negative root is behind the launch site.
page 6 of 7
C) Just as the balloon is released a bird flies past 20.0 m away and 20.0 m higher than the
roof. Will the balloon go above or below the bird? Justify your answer.
The bird is located at (20, 20). Let’s see if the balloon goes through this point.
y(x) = (tan θo)x – [(g/2cos2 θo)/vo2]x2
y(20) = (tan 45°)(20.0 m) – [(g/2cos2 45°)/(20.0 m/s)2](20.0 m)2
y(20) = (1)(20.0 m) – [(9.81 m/s2)/(400 m2/s2)](400 m2)
y(20) = 20.0 m – 9.81 m
y(20) = 20.0 m – 9.81 m = 10.19 m
y(20) = 10.2 m
No, the bird is not in danger. The balloon passes (20 m – 10.2 m) = 9.8 m below it.
page 7 of 7

Download Report