Projectile Motion: Calculating Time, Speed, and Velocity of a Thrown Dart

I can only tell you that if you follow the guideline you quoted, then you are going to end up not helping people who need it. There is no algorithm that tells you what is "too much" help. It's up to you.What do you need to know??In summary, the conversation discusses a problem involving projectile motion, specifically the calculation of the time of flight, initial speed, and velocity of a dart thrown by a player. The conversation also mentions the use of equations of motion and the importance of showing one's work in seeking help with a problem. The participants also discuss the issue of giving too much help and the guidelines for providing assistance on homework problems.
  • #1
misogynisticfeminist
370
0
I've got a question on projectile motion and don't know where to start.

A dart player stands 3m from the wall on which the board hangs and throws a dart which leavs his hand with a horizontal velocity at a point 1.8m above the ground. The dart strikes the board at a point 1.5 m from the ground. Calculate

a) The time of flight of the dart
b) The initial speed of the dart
c) The velocity of the dart when it hits the board.

The problem is at a) and i think once i know a) i should have no problems doing the rest. Thanks alot.
 
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  • #2
vertically the dart drop 0.3 meter... you know
[tex] x(t) = x_{y0}+v_{yo} t + 1/2 g t^2 [/tex]
what is the time...
 
  • #3
What are your ideas to solving the problem...??What do you need to know??


Daniel.
 
  • #4
There are 2 components of the motion to consider. The horizontal motion which is at constant speed (neglecting air resistance) and the vertical motion which undergoes a constant downward acceleration due to gravity.

Consider a Cartesian coordinate system (an [itex](x,y)[/itex] plane) where the dart starts at the origin and its position at time [itex]t[/itex] is [itex](x_t,y_t)[/itex]. Let [itex]v[/itex] be the initial horizontal velocity when the dart is released.

Then the equations of motion for the dart are :

[tex]x_t = vt[/tex]

[tex]y_t = -\frac{1}{2}gt^2[/tex]

(Can you see why ?)

Now figure out what [itex]y_t[/tex] should be at the point when the dart hits the board, and solve for [tex]t[/tex] using the second equation to get the first answer.
 
  • #5
My ideas so far are, to find an equation which has no variable, time,

[tex] V_y=V_o_y - gt [/tex]

But this would not be useful as I don't noe the initial and final velocities in the y-axis. But using displacement, which I have a lot of information on, I would get,

[tex] y= V_o_y t - \frac {1}{2} gt^2 [/tex]

just sub ing in y= 0.3, i would have 2 unknowns which is [tex] t [/tex] and [tex] V_o_y [/tex] and I can't seem to find another equation which contains these 2 unknowns as well.
 
  • #6
It would have been nice and useful to him,had he figured it out all by himself and not have his problem solved by someone else..;

Daniel.
 
  • #7
the dart is shooting horizontally, what do you think the [tex] V_{oy} [/tex] is

curious3141:
our spirit here is not to give the answer

did you ready the sticky, if don't, read it here
Read This Before Posting
I would like to say that if you are seeking help with problems that are giving you some heartburn, then you have come to the right place!

I would also like to say that if you are looking for someone to do your homework for you, then you have come to the wrong place.

Please, for your own sake, try your homework before you come here, and show us where you got stuck. Speaking for myself, I already know that I know how to do your problem. Me doing it for you accomplishes nothing.

So, post away--and show your work!
__________________



edit:
the quote above is a mistake... here you go
Participants:

Please do not hand out answers to homework problems. You should emphasize technique and methodology, but always leave at least some of the work to the student. Generally, the student must show some of his/her work, up to the point where he/she becomes stuck, before anyone should provide help.
 
Last edited:
  • #8
Yes, curious i do understand why it is [tex] x= V_x t [/tex] but not really [tex] y= -\frac{1}{2} gt [/tex]. Why do we assume [tex] V_o_y [/tex] to be 0? Doesn't it leave that guy's hand at a certain starting velocity?

edits: to vincent: Hmmm, ok, i did not read the part where it said "horizontal carefully". Lol i understand the rules in homework help. In fact, I've been trying to solve this for the past hour, and can't get started at all.

Thanks for the help, I should be able to handle the rest...

: )
 
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  • #9
dextercioby said:
It would have been nice and useful to him,had he figured it out all by himself and not have his problem solved by someone else..;

Daniel.
Ummm...I hardly think I gave away the answer. I explained the reasoning, gave basic equations that can be found in any introductory text, and let him do the rest. If I had plugged in the values and given the answer, then I can understand your bellyaching, but this was hardly the case.
 
  • #10
gaving out the equation has no different with plug in the number for him, you should let him think what formulas should he use... read my qoute above...
or here
Participants:

Please do not hand out answers to homework problems. You should emphasize technique and methodology, but always leave at least some of the work to the student. Generally, the student must show some of his/her work, up to the point where he/she becomes stuck, before anyone should provide help.
 
  • #11
vincentchan said:
gaving out the equation has no different with plug in the number for him, you should let him think what formulas should he use... read my qoute above...
or here

You gave essentially the same formula I did (except yours is illucid) AND you gave away what value to use for the drop in height (even though computing that would be trivial). In essence, you "gave away" more than I did.

I reckon I taught the orig. poster more about how to approach the problem because I explained the principles at work (no air resistance, gravity). You did not.

Look, arguing like this is pointless. It's a matter of personal judgement what "helping too much" is anyway. The only people I am against helping are the ones that expect to be spoonfed. This poster was obviously not one of those, because he even said he could do the rest of the parts himself.
 

What is projectile motion?

Projectile motion refers to the movement of an object through the air or space under the influence of gravity. It follows a curved path called a parabola.

What factors affect projectile motion?

The factors that affect projectile motion include the initial velocity, angle of launch, air resistance, and the force of gravity. These factors can impact the distance, height, and time of flight of the projectile.

How is the maximum height of a projectile calculated?

The maximum height of a projectile can be calculated using the formula h = (v sin θ)^2 / 2g, where h is the maximum height, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

What is the difference between horizontal and vertical components of projectile motion?

The horizontal component of projectile motion refers to the motion of the projectile along the x-axis, while the vertical component refers to the motion along the y-axis. The horizontal component remains constant while the vertical component is affected by gravity.

How can projectile motion be applied in real life?

Projectile motion has many practical applications, such as in sports like basketball, where players use the principles of projectile motion to make successful shots. It is also used in engineering and physics to design and analyze the trajectory of objects like rockets and projectiles.

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