Help Me Find Answers to My Stolen Homework Test Due at 11 pm CT

[Keem]

I went in for tutoring on an online homework test that is due tonight and I realize that it was stolen off my desk while I went to sleep. Its a long story but I don't have time to explain. If someone can please help me with these questions. Only 2 and there are six. I have the other 4 correct
first one is
A golfer can hit a golf ball a horizontal distance of over 300 m on a good drive.
What maximum height will a 318.3m drive reach if it is launched at an angle of 15 degrees to the ground? Answer in M

A shell is fired from the ground with an initial speed of 1.60x10^3 m/s at an initial angle of 41 degrees to the horizontal

a)Find the shells horizontal range
258418.579

b)How long is the shell in motion? Answer in s.
I don't know this one. If someone can give me quick help I will deeply appreciate it. The homework is due 11 pm CT

 

[Keem]

as far as the work goes I was given an equation and told to work backwards but I still can't figure it
for #2
the range equation
R=(VO^2sin(2theta)/lgl
318.3=(VO^2sin(2*15)/9.81
318.3*9.81/2sin(30)=VO^2
sq root of 6238.68= Sq Root VO^2

 

[mezarashi]

Part b can be answered many different ways. But the easiest would be to look at your answer from part a.

You know the shell's horizontal range. d = vt. If you know its velocity in the horizontal direction you can most definitely tell how long it would've took to get there.

 

[stmoe]

... you know how far it went.. you know how fast it was going when it started moving... you know that it didn't slow down in the horizontal...
.. the identity for sin(2theta) is 2 sin(theta) cos(theta) ...
I did a proof of this my sophomore year (of HS ) ... of the range equation, its not really that difficult.. lke.. range = veloc in x * time .. from that you can get it in like 4/5 steps .. less if you can visualize it really quickly

... just out of curiosity... if its an online homework test... and someone stole it from your desk .. how are you posting to an online forum about it ? ... that's like saying 'help, someone stole my keyboard' or better yet.. the monitor -_-

 

[Keem]

... you know how far it went.. you know how fast it was going when it started moving... you know that it didn't slow down in the horizontal...
.. the identity for sin(2theta) is 2 sin(theta) cos(theta) ...
I did a proof of this my sophomore year (of HS ) ... of the range equation, its not really that difficult.. lke.. range = veloc in x * time .. from that you can get it in like 4/5 steps .. less if you can visualize it really quickly

... just out of curiosity... if its an online homework test... and someone stole it from your desk .. how are you posting to an online forum about it ? ... that's like saying 'help, someone stole my keyboard' or better yet.. the monitor -_-

I know it sounds strange but its a long story. I didn't realize it till the last second but to prove I did the work I have the answers to the questions. I went in for tutoring today and everything was on that one sheet. I went to more tutoring after school and I realize that I have a paper that isn't mine. Either someone picked it up off my desk and swapped it or someone just picked up the wrong paper because I have a paper I never seen before.

This print-out should have 6 questions.
Multiple-choice questions may continue on
the next column or page { ¯nd all choices
before answering. The due time is Central
time.

001 (part 1 of 1) 10 points
An autographed baseball rolls o® of a 0.53 m
high desk and strikes the °oor 0.39 m away
from the desk.

The acceleration of gravity is 9:81 m=s2 :
How fast was it rolling on the desk before it
fell o®? Answer in units of m=s.
002 (part 1 of 1) 10 points

1.186

A golfer can hit a golf ball a horizontal dis-
tance of over 300 m on a good drive.
What maximum height will a 318.3 m drive
reach if it is launched at an angle of 15:0± to
the ground? Answer in units of m.
003 (part 1 of 2) 10 points

78.985 WRONG

A ferry is crossing a river. The ferry is headed
due north with a speed of 1.7 m/s relative to
the water and the river's velocity is 2.1 m/s
to the east.
a) What is magnitude of the boat's velocity
relative to Earth? Answer in units of m=s.
004 (part 2 of 2) 10 points

2.702

b) Find the direction in which the ferry is
moving (measured from due east, with coun-
terclockwise positive). Answer in units of ±.
005 (part 1 of 2) 10 points

38.991

A shell is ¯red from the ground with an initial
speed of 1:60 £ 103 m/s (approximately ¯ve
times the speed of sound) at an initial angle
of 41:0± to the horizontal.
The acceleration of gravity is 9:81 m=s2 :
a) Neglecting air resistance, ¯nd the shell's
horizontal range. Answer in units of m.
006 (part 2 of 2) 10 points

258419

b) How long is the shell in motion? Answer in
units of s.

 

[Keem]

A golfer can hit a golf ball a horizontal dis-
tance of over 300 m on a good drive.
What maximum height will a 318.3 m drive
reach if it is launched at an angle of 15:0± to
the ground? Answer in units of m.
003 (part 1 of 2) 10 points
78.985 WRONG

A shell is ¯red from the ground with an initial
speed of 1:60 £ 103 m/s (approximately ¯ve
times the speed of sound) at an initial angle
of 41:0± to the horizontal.
The acceleration of gravity is 9:81 m=s2 :
a) Neglecting air resistance, ¯nd the shell's
horizontal range. Answer in units of m.
006 (part 2 of 2) 10 points
258419
b) How long is the shell in motion? Answer in
units of s.

if range = v in x * time then this is how i figure it out
258418.579=(sin41)1.60*10^3*t
258418.579=1049.694*t
258418.579/1049.694=t
246.1845=t

which is incorrect

 

[mezarashi]

if range = v in x * time then this is how i figure it out
258418.579=(sin41)1.60*10^3*t

Why did you use $$v_x = v\sin41^o$$?

 

[Keem]

Why did you use $$v_x = v\sin41$$?

I thought that would help me determine the v of x since the I have VO and the angle.

 

[mezarashi]

A shell is fired from the ground with an initial speed of 1.60x10^3 m/s at an initial angle of 41 degrees to the horizontal

Referring to the original question, the initial angle is 41 degrees to the horizontal. Now using your equation Vx = Vsin(angle), see if the following situation makes sense.

The shell is now fired at 90 degrees to the horizontal (i.e. directly upwards), the velocity in the x direction is Vx = Vsin(90) = V.

 

[Keem]

Referring to the original question, the initial angle is 41 degrees to the horizontal. Now using your equation Vx = Vsin(angle), see if the following situation makes sense.

The shell is now fired at 90 degrees to the horizontal (i.e. directly upwards), the velocity in the x direction is Vx = Vsin(90) = V.

I'm not sure. My instructor told us to use the angle and compare it to what side we were trying to find. The known velocity is 1.6*10^3 m/s at 41 degrees to the horizontal. so inorder to find the v of the x I would have to take the sin of 41 and multiply it by 1.6*10^3.
Understand I'm trying my hardest to figure this out

 

[Keem]

cann someone please explain this to me from beginning at least. I have 15 minutes left. I can't figure this out.
for no 2
I have the g, the range, the angle,
I don't know the time, the change in x
What I need to know is the max height (I don't know if its the change in y or what)

For letter b I have the range, the angle, VO (the hypotenuse)
What I don't have is the time
What I need to know is time

 

[Keem]

now I officially have a 64 as a test grade.
Thanks a lot you guys! you explained everything.

 

[mezarashi]

I'm not sure. My instructor told us to use the angle and compare it to what side we were trying to find. The known velocity is 1.6*10^3 m/s at 41 degrees to the horizontal. so inorder to find the v of the x I would have to take the sin of 41 and multiply it by 1.6*10^3.
Understand I'm trying my hardest to figure this out

Which is what I'm worried about. You should instead be trying to understand what your instructor is teaching you rather than following set rules. If you draw a right triangle of the velocity vector, don't you see that the x component is a Vcosine(angle) rather than sine?

now I officially have a 64 as a test grade.
Thanks a lot you guys! you explained everything.

I take that as sarcasm, and I don't think it would be proper to capitalize on this forum in order to make better grades. It may have been your responsibility to understand concepts prior to a test or quiz. The forum offers voluntary assistance not a guarantee.

 

[Keem]

Which is what I'm worried about. You should instead be trying to understand what your instructor is teaching you rather than following set rules. If you draw a right triangle of the velocity vector, don't you see that the x component is a Vcosine(angle) rather than sine?



I take that as sarcasm, and I don't think it would be proper to capitalize on this forum in order to make better grades. It may have been your responsibility to understand concepts prior to a test or quiz. The forum offers voluntary assistance not a guarantee.

yes it was my responsibility. I know that but still an equation could've really helped. I tried my hardest and obviously didn't understand. I wasn't asing for an answer. Just an explanation.

 

[mezarashi]

So that you'll be more prepared next time, if there is one, you should practice on more questions. A trick to introductory 2-D particle kinematics is that you have the 5 kinematics equations:

a = v / t ... 1
d = 1/2 (v2 + v1)t ... 2
d = v1t + 1/2 a(t)^2 ... 3
d = v2t + 1/2 a(t)^2 ... 4
v2^2 = v1^2 + 2ad ... 5

Except for (1), if you know any three variables out of five (a,t, vi,vf,d), you can always find the fourth. Then with the fourth you can find the fifth. When angles are involved, you will need to solve 2 equations simultaneously.

 

[Keem]

thanks for trying you all. I'll admit my last comment was a little immature because I was frustrated but there should be extra credit. I'm taking this senior year so its not that easy for me and science is my weakest class right now. But I'm over it.