Projectile Motion Tossing a Bag from Dayton's to IDS

[~christina~]

Projectile motion? with a bag being thrown up to another building higher than the 1st

Homework Statement

The world renowned Forensic Scientist, Dr. Hayu Iamasleuth, sits at his desk poring over notes on a case in which he is a consultant for the defence attorney. According to the crime scene report, a thief allegedly that is located next to the Dayton's. To escape the pursuing guard the alleged thief took th express elevator to the roof of the Dayton's where she threw the money bag to a waiting accomplice on the roof of the IDS crystal court which is just EAST of Dayton's. The IDS crystal court and Dayton's are separated by the Nicollet Mall. There was an eye witness to the crime, Ms. Lajoyce Hilchrist who is a criminalist who earned a reputation for her testimony cutting through the fog of doubt with hard science. No one else witnessed the crime and the security camera did not catch the thief. He examines her testimony carefully. According to her the defendent arrived on the roof no more than 2 seconds before the guard did.
The money bag was tossed from the edge of the Dalton's roof with a force just great enough for it to reach the accomplice on the IDS roof. Then the alleged thief lit a cigarette and leaned against a air condition unit as if she had been there for some time. The IDS tower is 250m high, Dayton is 100m high, and the Nicollet Mall is 20m wide.

a) With what initial velocity (magnitude and direction) was the money bag tossed if Ms. Hilchrist statement is accurate?

b) What was the time of flight of the money bag?

c) sketch graphs of the x and y compionent of the displacement, velocity and acceleration of the bag as functions of time for the interval starting when the bag leaves Dayton's and ending when it reaches the IDS roof (6 graphs)

d) Can Hay use the above findings to discredit the witness? Use relevant physics principle(s) to quantitatively and qualitatively explain your response.

(Note: Jose Canseco throws a baseball with an initial velocity of 45m/s; John Elway throws a football with an initial velocity of 31m/s)

What I thought was the relevant information I bolded.

Homework Equations

If this a projectile motion question wouldn't I use the kinematic equations?
However the bag is thrown with a velocity plus it is being affected by gravity...would this be a normal projectile motion question?
I think so but not exactly since it's thrown up if I'm not incorrect and I'm not sure how that would work since I don't have a angle that it was thrown from..

The Attempt at a Solution

Well I'm not sure where to start but I do know that...

h Dalton= 100m
h IDS crystal court = 250m
d or length of the Nicollet mall= 20m wide

vi of throw= ?
t of moneybag flight= ?

I don't know why I'm given the velocity of a baseball and football at the end but i assume that it would provide me with a reasonable value to compare with when I find the initial velocity of the throw..

I dont' know where to start unfortunately with this problem

Thanks

 

[~christina~]

well now I'm thinking that acelleration would be positive 9.8m/s^2 since if I consider the origin at the Dalton's building and the bag goes up then it should be posititve if I consider the downward direction negative but wouldn't the bag's aceleration be negative for a time if it goes up and then down a little ? but I don't know if the bag just reaches the max of the parabola then the accomplice grabs the bag or does it past the top of the other building then comes down where the accomplice catches the bag.

I need help on starting this since I haven't encountered a question quite like this before.

 

[hotvette]

Part of the difficulty is sorting through all of the text to extract the relevant information.

You are correct that this is a classic projectile problem. You are given a horizontal distance and vertical distance the object must travel. Write the kinematic equations for a projectile and go from there. As you implied, it is important to label your origin and axes so everything can be kept straight.

 

[~christina~]

Woudn't I have to know the angle or not?


a) With what initial velocity (magnitude and direction) was the money bag tossed if Ms. Hilchrist statement is accurate?
If I don't have the time of flight and just the distances traveled and I don't know a angle or anything how am I supposed to find the initial velocity?
I am perplexed by this

b) What was the time of flight of the money bag?
I can't find this unless I find the other part??


~I really don't know what kinematic equation I'm supposed to use since I don't have a bunch of information except for the distances traveled...would the final velocity be zero??


HELP

 

[learningphysics]

The main trick is that the maximum height that the bag reaches is 250-100 = 150m

With this information it is possible to find the velocity and angle...

 

[~christina~]

Well I found this...

h/R= tan theta/4

h= 150m
R= 20m

150m/20m= tan theta/4
tan theta= 30
tan^-1(30)= 88.09

so the angle = 88.09 deg

using that to get the initial velocity...

vo=\sqrt{} Rg/ sin 2theta)

vo=\sqrt{} (20m*9.8m/s^2)/ sin 2*88.09

vo= \sqrt{} 2941.96252

vo= 54.240m/s


b.) To find the time of flight of the money bag..I'm not quite sure which equation to use but would it be:

y= Voy*t + 1/2*g*t^2?

where y would equal 150m
and Voy would equal 54.240 m/s that I found then would I use the quadradic formula to find the t for the moneybag?


Thanks =D

 

[learningphysics]

Voy = Vosin(theta) = 54.24sin(88.09)

yes, you'd use the quadratic formula to find time...

everything you did looks right. I'd advise to try and work out the problem from the basic kinematics formulas also and see if you get the same answers... it's always good to try and solve these from the basic formulas because it improves your understanding of the fundamentals...

 

[~christina~]

Need to finish this by today...Please!

b) I plugged in the Voy (Voy= 54.21m/s) into the equation to find the time of the money bag.

dy= Voy(t) + 1/2gt^2

dy= 150m
g= 9.8m/s^2
Voy= 54.21m/s

150m= 54.21m/s(t) + 4.9m/s^2(t)^2
4.9m/s^2 (t)^2 + 54.21m/s (t)- 150 = 0

quad formula:
(-54.21 +/- \sqrt{} (54.21)^2 - 4(4.9)*(-150))/ 2(4.9)

t= 2.29s

~This seems a really small time..is it alright?


c.) graphing...if they say to sketch do I have to go and actually plot everything??

For the equations I got:

displacement vs time:
x component: dx(t)= 1.80t
y component: dy(t)= 54.21m/s(t) -4.9t^2

velocity vs time:
Vox= Vo Cos(theta)
Vo= 54.21m/s
theta= 88.09 deg

Vox= 54.24Cos(88.09)
Vox= 1.80m/s

x component: vx(t)= 1.80m/s

y component:
Voy= 54.21m/s
a= -9.8m/s

Vyf= Voy + ax(t)

vy(t)= 54.21m/s - 9.8m/s^2(t)

Acceleration vs time:
x component: ax(t)= 0

y component: ay(t)= -9.8m/s^2


_________________________________
d.) If he can discredit the witness..

~I say yes b/c the info given in the problem states in a note that (Jose Canseco throws a baseball with an initial velocity of 45m/s; John Elway throws a football with an initial velocity of 31m/s)

The velocity found (vi= 54.24m/s) is higher than what a baseball player can throw a baseball that I assume would logically weigh more than a bag filled with money.

It would if the person threw it at that angle with that velocity reach the target though...
but I don't think a human can throw a money bag at a higher velocity than a baseball player who throws baseballs faster than the normal person.



Is what I did fine??

I need to finish this by the end of today...Please

Thank You Learning Physics

 

[learningphysics]

No, Voy is not 54.21m/s... it is 54.21*sin(88.09)

 

[~christina~]

Voy = Vosin(theta) = 54.24sin(88.09)

=>you typed that and that was where I got Voy...which was 54.21m/s

Why Is that??

the initial velocity I found was Vo= 54.24m/s
and I plugged that into the eqzn for the Voy..

 

[learningphysics]

=>you typed that and that was where I got Voy...which was 54.21m/s

Why Is that??

the initial velocity I found was Vo= 54.24m/s
and I plugged that into the eqzn for the Voy..

Oh... I'm sorry... You're right... I didn't notice that it was different...

You're right Voy =54.21. I apologize. I got confused... coz it's so close to 90 degrees.

 

[learningphysics]

Ah... sorry I didn't point this out earlier... the best way to get the time is:

horizontal distance/horitonztal velocity = 20m/54.24cos(88.09) = 20m/1.8 = 11s

The quadratic should also give the same time... I think the reason you didn't get that time is because you used a = 9.8 instead of a = -9.8

 

[~christina~]

dy= Voy(t) + 1/2gt^2

dy= 150m
g= 9.8m/s^2
Voy= 54.21m/s

150m= 54.21m/s(t) + 4.9m/s^2(t)^2
4.9m/s^2 (t)^2 + 54.21m/s (t)- 150 = 0

quad formula:
(-54.21 +/- )/ 2(4.9)

t= 2.29s

but when I used the negative I got a imaginary number since I can't get the radical of a negative ...why is that??

 

[learningphysics]

I think I might see the problem... the R needs to be the range of the projectile... in this case R is 40m not 20... in other words the total distance it "would have gone" if the object kept going and landed at the same height it left from...

So 150/40 = tan(theta)/4

theta = 86.186degrees

so that will change your vo.