Help With Physics

[golferguy12]

Hey Guys-

I got this physics class that is really giving me trouble. I was wondering if any of you could give me a hand on some of my problems. I really appreciate it.

1. A mountain climber jumps a 1.0 m wide crevasse by leaping horizontally with a speed of 7.3 m/s.
   a)If the climber’s direction of motion on landing is -45°, what is the height difference between the two sides of the crevasse?
   (b) Where does the climber land?
   (BLANK) m beyond the far edge

2. In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower height is h = 7.80 m, and that the bulls-eye’s horizontal distance is d = 3.9 m from the launch point. If the pumpkin is thrown horizontally, what is the launch speed needed to hit the bulls-eye? (Neglect air resistance.)

3. Fairgoers ride a Ferris wheel with a radius of 5.00 m, the bottom of which is 1.75 m above the ground. The wheel completes one revolution every 36.0 s.
   a) What is the average speed of the rider?
   b)If a rider accidentally drops a stuffed animal at the top of the wheel, where does it land relative to the base of the ride? (Neglect air resistance.)

4. In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14 m/s at an angle of 20° above the horizontal.
   (a) How long does it take for the ball to reach the wall if it is 5.7 m away?
   (b) How high is the ball when it hits the wall?

5. A golf ball is struck with a five iron on level ground. It lands 100.0 m away 3.40 s later. What was the magnitude and direction of the initial velocity? (Neglect air resistance.)

6. A bomber is flying horizontally over level terrain, with a speed of 255 m/s relative to the ground, at an altitude of 3100 m. Neglect the effects of air resistance.
   (a) How far will a bomb travel horizontally between its release from the plane and its impact on the ground?
   (b) If the plane maintains its original course and speed, where will it be when the bomb hits the ground?
   behind the bomb
   ahead of the bomb
   directly above the bomb
   (c) At what angle from the vertical (less than 90°) should the telescopic bombsight be set so that the bomb will hit the target seen in the sight at the time of release?

[kelleyb]

So… you want us to do your homework for you? I can almost guarantee you if you try to answer those questions on your own and then post both Q and A on here and ask for feedback you’ll get some good advice. But don’t expect to be spoon fed your answers.

[Regular_Gonzalez]

You would probably be better off asking your teacher for help, or hiring a private tutor if you can afford it.

Here is my attempt at question 1:

1(a)

The angle at landing is 45 degrees, so I assume that vertical and horizontal velocity are equal at this point (7.3 m/s).

Acceleration due to gravity = 9.8m/s

Time = 7.3/9.8 = 0.75 sec

Average vertical velocity = 7.3/2

Height difference = avge vertical velocity x time = (7.3/2)x0.75sec = 2.7m

(b)
Jump distance = 7.3 m/s x 0.75 sec = 5.4 m

[lixy]

http://www.catb.org/~esr/faqs/smart-questions.html

[Scrotus]

If you have to ask T-Nation this, you are fucked come test time.

[golferguy12]

Alright, thanks for the responses. The thing about this class is that it’s completely graded off of homework, which is turned in online. You get three guesses per question, and at the end all of your homework is added up and that is your grade. This may seem easy, but there is homework every night and I spend 4-5 hours per night on it and still get less than 50% right. The entire class is averaging around that.

The teacher said its supposed to be hard, thats why we get three guess per question, and even the tutors I went to could not figure it out. I would go in early, but I work from 6-11 am (class 12-2 M-Thurs.) and then I use the rest of my day to actually do the homework or practice golf (collegiate golfer). I was just hoping somebody could help me get some points and maybe change my mind about dropping, but I guess not. Understandable, thanks for the tries.

[blazindave]

[quote]golferguy12 wrote:
2. In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower height is h = 7.80 m, and that the bulls-eye’s horizontal distance is d = 3.9 m from the launch point. If the pumpkin is thrown horizontally, what is the launch speed needed to hit the bulls-eye? (Neglect air resistance.)
[/quote]

Known factors:

-Gravity: -9.8 meters per second per second
-Distance of floor from tower: -7.8 meters
-Distance of bull’s eye (assuming it is flat on the ground and level with floor): 3.9 meters
-Jack o lantern is thrown in a perfectly horizontal fashion.
-Air resistance is neglected.

Equations which might help:
S = (vit) - (.5a*t^2)
displacement = (initial velocity * time) - (.5 * acceleration * time squared)

Let a = acceleration = gravity = g

We first need to know how long it takes for the jack o lantern to hit the ground:

S = 7.8 meters
t = unknown
a = 9.8m/s^2
initial velocity = vi = 0 m/s

7.8 = 0 - .5*(-9.8)*t^2
7.8 = - (-4.9)t^2 or 15.6 = -(-9.8)t^2
7.8 = 4.9t^2 or 15.6 = 9.8t^2
7.8/4.9 = t^2 or 15.6/9.8 = t^2

=> 1.59 = t^2
Root of 1.59 = t =1.26 seconds

Now let’s look at the horizontal components:

Distance = 3.9 meters
-So we need a displacement of 3.9 meters horizontally
So let S = 3.9 meters

-Launch speed = initial velocity = vi = unknown

-Time it takes is dependent on how long it takes the ball to hit the ground. Since we know that the time is 1.26 seconds, we let t = 1.26.

-Since there is no horizontal acceleration (horizontal gravity doesn’t exist), acceleration ( =a ) is 0.

Using the same equation as before:
S = (vit) - (.5a*t^2)

3.9 = (vi * 1.26) - (.5* 0 *1.26^2)

We know that anything times zero is zero, so:

3.9 = (vi * 1.26) - 0

3.9 = vi * 1.26

3.9/1.26 = vi = 3.1 meters per second

Let’s check with symbols/units.

S = meter
vi = meter per second
t = second

meter = (meter/second) * second
meter/second = meter/ second
so it checks out.

Good luck with the rest.

[CJK]

[quote]blazindave wrote:
golferguy12 wrote:
2. In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower height is h = 7.80 m, and that the bulls-eye’s horizontal distance is d = 3.9 m from the launch point. If the pumpkin is thrown horizontally, what is the launch speed needed to hit the bulls-eye? (Neglect air resistance.)

Known factors:

-Gravity: -9.8 meters per second per second
-Distance of floor from tower: -7.8 meters
-Distance of bull’s eye (assuming it is flat on the ground and level with floor): 3.9 meters
-Jack o lantern is thrown in a perfectly horizontal fashion.
-Air resistance is neglected.

Equations which might help:
S = (vit) - (.5a*t^2)
displacement = (initial velocity * time) - (.5 * acceleration * time squared)

Let a = acceleration = gravity = g

We first need to know how long it takes for the jack o lantern to hit the ground:

S = 7.8 meters
t = unknown
a = 9.8m/s^2
initial velocity = vi = 0 m/s

7.8 = 0 - .5*(-9.8)*t^2
7.8 = - (-4.9)t^2 or 15.6 = -(-9.8)t^2
7.8 = 4.9t^2 or 15.6 = 9.8t^2
7.8/4.9 = t^2 or 15.6/9.8 = t^2

=> 1.59 = t^2
Root of 1.59 = t =1.26 seconds

Now let’s look at the horizontal components:

Distance = 3.9 meters
-So we need a displacement of 3.9 meters horizontally
So let S = 3.9 meters

-Launch speed = initial velocity = vi = unknown

-Time it takes is dependent on how long it takes the ball to hit the ground. Since we know that the time is 1.26 seconds, we let t = 1.26.

-Since there is no horizontal acceleration (horizontal gravity doesn’t exist), acceleration ( =a ) is 0.

Using the same equation as before:
S = (vit) - (.5a*t^2)

3.9 = (vi * 1.26) - (.5* 0 *1.26^2)

We know that anything times zero is zero, so:

3.9 = (vi * 1.26) - 0

3.9 = vi * 1.26

3.9/1.26 = vi = 3.1 meters per second

Let’s check with symbols/units.

S = meter
vi = meter per second
t = second

meter = (meter/second) * second
meter/second = meter/ second
so it checks out.

Good luck with the rest.
[/quote]

My mind just exploded.

[Gael]

5. Vx=d/t=100/3.4=29.4
   Vy=at/2=4.8*3.4=16.7
   Vmag=root(Vx^2+Vy^2)=33.8 m/s
   Angle=arctan(16.7/29.4) = 29 degrees

[Gael]

4. Vx=14cos20=13.1 m/s
   t=d/v=5.7/13.1=.43 seconds
   d=14sin20-.5at^2= 3.9 meters

[Gael]

6. dy=4.9t^2=3100 … t=root(3100/4.9)=25.2 s
   dx=255*25.2=6414 m

directly above

arctan(3100/6414) = 25.8 degrees

[HardcoreHorn]

Don’t you have a Physics book to go with the class? Just use the kinematics equations and plug in the given values to solve for the unknown values. That’s all these problems are.

[dasboi]

[quote]golferguy12 wrote:

1. A mountain climber jumps a 1.0 m wide crevasse by leaping horizontally with a speed of 7.3 m/s.
   a)If the climber’s direction of motion on landing is -45°, what is the height difference between the two sides of the crevasse?
   (b) Where does the climber land?
   (BLANK) m beyond the far edge

[/quote]

a) horizontal velocity = 7.3m/s = constant
initial vertical velocity = final vertical velocity =7.3tan45
=7.3m/s
vertical distance travelled = 0.5vt = 0.5v(square)/a
= 0.57.3*7.3/9.81
= 2.716m

b) time taken for fall = v/a= 0.744s
horizontal distance travelled = 7.3*0.744 = 5.43m
therefore, climber lands 4.43m away from the edge.

[dasboi]

[quote]golferguy12 wrote:

2. In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower height is h = 7.80 m, and that the bulls-eye’s horizontal distance is d = 3.9 m from the launch point. If the pumpkin is thrown horizontally, what is the launch speed needed to hit the bulls-eye? (Neglect air resistance.)

[/quote]

time taken to drop 7.8m = 7.8/9.81 = 0.795s
therefore, launch speed = 3.9/0.795 = 4.91m/s

[dasboi]

[quote]golferguy12 wrote:

3. Fairgoers ride a Ferris wheel with a radius of 5.00 m, the bottom of which is 1.75 m above the ground. The wheel completes one revolution every 36.0 s.
   a) What is the average speed of the rider?
   b)If a rider accidentally drops a stuffed animal at the top of the wheel, where does it land relative to the base of the ride? (Neglect air resistance.)

[/quote]

a) circumfrence of wheel = 3.142*10 = 31.42m
average speed = 31.42/36 = 0.872m/s

b) acceleration = 9.81+ 0.152 (it is circular motion. so take note of the added acceleration) = 9.962
vertical distance dropped = 11.75m
time taken = sqrt(211.75/9.962) = 1.536s
horizontal distance travelled = 0.8721.536 = 1.34m

[dasboi]

[quote]golferguy12 wrote:

4. In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14 m/s at an angle of 20° above the horizontal.
   (a) How long does it take for the ball to reach the wall if it is 5.7 m away?
   (b) How high is the ball when it hits the wall?

[/quote]

a) horizontal velocity = 14cos20 = 13.16m/s
time taken = 5.7/13.16 = 0.433s

b) initial vertical velocity = 14sin20 = 4.78m/s
final vertical velocity = 4.78 - 9.810.433 = 0.532m/s
max vertical distance ball could travel if undisturbed = 0.54.784.78/9.81 = 1.16m
height ball hits the wall = 1.16 - 0.50.532*0.532/9.81 = 1.146m

[dasboi]

[quote]golferguy12 wrote:

5. A golf ball is struck with a five iron on level ground. It lands 100.0 m away 3.40 s later. What was the magnitude and direction of the initial velocity? (Neglect air resistance.)

[/quote]

max vertical velocity = 9.813.4 = 33.354m/s
max height = 0.533.354*1.7 = 28.35m
direction of inital velocity = inverse tangent(28.35/50) = 29.55 =30 degrees
initial velocity = 33.354/sin30 = 66.708m/s

[dasboi]

[quote]golferguy12 wrote:

6. A bomber is flying horizontally over level terrain, with a speed of 255 m/s relative to the ground, at an altitude of 3100 m. Neglect the effects of air resistance.
   (a) How far will a bomb travel horizontally between its release from the plane and its impact on the ground?
   (b) If the plane maintains its original course and speed, where will it be when the bomb hits the ground?
   behind the bomb
   ahead of the bomb
   directly above the bomb
   (c) At what angle from the vertical (less than 90°) should the telescopic bombsight be set so that the bomb will hit the target seen in the sight at the time of release?
   [/quote]

a) time taken to drop 3.1km = sqrt(23100/9.81) = 25s
distance travelled = 25525= 6375m

b) directly above

c) angle = inverse tangent(6375/3100) = 64degrees

[dasboi]

is it really that hard? i’m guessing your not doing a science or math based major if thats the case.

just get loads of practice man.

[HoratioSandoval]

[quote]golferguy12 wrote:
Alright, thanks for the responses. The thing about this class is that it’s completely graded off of homework, which is turned in online. You get three guesses per question, and at the end all of your homework is added up and that is your grade. This may seem easy, but there is homework every night and I spend 4-5 hours per night on it and still get less than 50% right. The entire class is averaging around that.

The teacher said its supposed to be hard, thats why we get three guess per question, and even the tutors I went to could not figure it out. I would go in early, but I work from 6-11 am (class 12-2 M-Thurs.) and then I use the rest of my day to actually do the homework or practice golf (collegiate golfer). I was just hoping somebody could help me get some points and maybe change my mind about dropping, but I guess not. Understandable, thanks for the tries.[/quote]

C’mon, what kind of retarded tutors do you have? This is all easy stuff.

It’s got to be your first or second week of class - are you already going to go down in a ball of flames?

If your major requires physics, it’s going to get wayyyyy harder than this. One class is not enough. I assume you’ve had calc so you’re not taking this to get a science w/a lab. What’s your major? Dropping the class will not help unless you radically change your major. Are you ready for that?