Projectile Motion So many Unknowns

[orangem]

Homework Statement

Need Help
I have been trying to solve this problem for a while now. I know the quadratic formula has to be used in order to solve this problem but I am having trouble getting started.

Question

A soccer ball is kicked from the ground level at an angle, such that the wall is hit at (point C) 1.9 seconds later. Point C (the wall) is 5.7m above the ground and the wall is 30m from where the ball is kicked. Ignore the effect of air resistance. Determine the magnitude and direction of initial velocity. How high the ball goes (point B, the highest point) the magnitude and direction of the velocity at the point of impact with the wall.

Unknown:
Voy-not given
Vox-not given
Vmag-not given
Angle -not given
Distance to point B-not given
Time to point B-not given
Time from point B to wall-not given
Distance from point B to wall not given

Known:
total time 1.9 sec
Height of wall where soccer ball hits 5.7m
Displacement -30m

ay=-9.8
ax=0
Vox=Vx

At Part B:
Vy = 0

Homework Equations

Vf=Vo+at
DeltaX=V(T)+ .5(at)
Vf^2=Vo^2=2(at)

ax2 + bx + c = 0

The Attempt at a Solution

I am not sure which formula to use since there are so many unknowns. I was thinking about using the distance formula just to part B and use the quad formula to solve to the two times but the distance to part B is not given.

Or should I use the third equation and say the final velocity is zero?

Is this how to get the Angle:
Triangle
30m, 5.7m, and X as the hypothenuse.
30^2+5.7^2=X^2

X= 30.5
Angle= arc Tan (5.7/30.5)= 10.75

 

[orangem]

Actually Never mind about the angle. I know that is completely wrong for this kind of problem.

 

[nasu]

For this kind of problem you need to write the equations of motion separately for the horizontal and vertical directions (x and y). Then see what is given for each direction and can be found.
Your equation are some sort of mixture between the two directions.
What kind of motion is the horizontal component?

 

[orangem]

I know they need to be broken up into two components. There is just not enough information to go from here
X:
Vox-?, Vfx-? total distance 30 m
vx = v0x, x = v0xt,
Y:
Voy-?, Vfy-?, Vy at point B-Zero, Point B-?, Point C- 5.7m
vy= v0y - gt, y = v0yt - (1/2)gt2.

Time 1.9sec

 

[collinsmark]

I know they need to be broken up into two components. There is just not enough information to go from here

There's enough information.

X:
Vox-?, Vfx-? total distance 30 m
vx = v0x, x = v0xt,

Okay, let's start with the x-component before moving to the y.

What are the forces involved in the x-direction? (Remember, you get to ignore air resistance.) Are there any? What does that tell you about the acceleration in the x-direction? And thusly, what does that tell you about the relationship between v_0x and v_fx?

By the way, before we move onto the y-direction, has your coursework introduced the concept of conservation of mechanical energy yet? If so, it will make the rest of this problem a little easier. If not, it's still solvable anyway. I'm just curious is all.

 

[azizlwl]

deleleting my post

 

[Yukoel]

Hello orangem,
Well the problem involves you to understand the following points:
(i) How does the horizontal component of velocity change with time?And by that nature how can we express the horizontal displacement of the ball w.r.t the time.The horizontal distance corresponding to a given time is given right?
(ii)How does the vertical component change w.r.t time? How does one express the vertical displacement in given time then?The vertical displacement and the time taken also given.Calculate the maximum height of the projectile.
Then after calculating the components the magnitude and the direction are asked to be calculated.
Note that in your fourth post you gave v(y)=0 at collision which may/may not be true.
Does this help?
regards
Yukoel

 

[orangem]

Ok x component
No air resistance. The initial velocity will be the same as the final velocity. And there is no acceleration. So as I previously stated Vfx=Vox. Does this mean the Vox=15.8 because displacement=VoxT. So Vx=15.8 and Vox= 15.8

 

[collinsmark]

Ok x component
No air resistance. The initial velocity will be the same as the final velocity. And there is no acceleration. So as I previously stated Vfx=Vox. Does this mean the Vox=15.8 because displacement=VoxT. So Vx=15.8 and Vox= 15.8

(Don't forget your units.)

Okay, that looks good so far. Now come up with an expression for v_yi as a function of θ_i. (Keep it in terms of θ_i for now, since you don't know what that is yet. Don't worry, you'll be solving for θ_i soon enough.)

 

[Yukoel]

Ok x component
No air resistance. The initial velocity will be the same as the final velocity. And there is no acceleration. So as I previously stated Vfx=Vox. Does this mean the Vox=15.8 because displacement=VoxT. So Vx=15.8 and Vox= 15.8

Sounds correct
Now for the y component?

regards
Yukoel

 

[orangem]

Sweet.
ok anyways vy = v0sinθ-at

 

[SammyS]

Ok x component
No air resistance. The initial velocity will be the same as the final velocity. And there is no acceleration. So as I previously stated Vfx=Vox. Does this mean the Vox=15.8 because displacement=VoxT. So Vx=15.8 and Vox= 15.8

Well, the horizontal component of the ball is 30m, & it covers this horizontal distance in 1.9 seconds, so (V₀)_x = 30/1.9 . Is that how you got 15.8 m/s ?

If there were no gravity, the ball would travel in a straight line, correct? Let's call that point, D.

But there is gravity, so the ball hits at point C. The distance from D down to C is how far the ball will drop in 1.9 seconds. (It's just like the "monkey shoot" if you're familiar with that.) That's one way to figure this out without too many equations. --- I don't much like looking through a bunch of equations to randomly pick out one that might work.

 

[Yukoel]

Sweet.
ok anyways vy = v0sinθ-at

The equation works no doubt with a=g but will that help you in getting the y component here?Think of an equation of kinematics which correlates the total displacement with initial velocity ,time and acceleration.You have the vertical displacement ,the acceleration and the time taken given so the vertical velocity (initial) is the only unknown.
Does this help?
regards
Yukoel

 

[orangem]

y=VoyT-.5*gt^2 but I do not know the vertical displacement

 

[orangem]

or Voy

 

[collinsmark]

y=VoyT-.5*gt^2 but I do not know the vertical displacement

From the problem statement:

"Point C (the wall) is 5.7m above the ground."​

So you know what 'y' is, you know what t is, you know what g is, so you should be able to solve for v_0y, right?

 

[orangem]

is it 12.31m/s

5.7=Voy (1.9) - .5(9.8)(1.9^2)
5.7=Voy (1.9) - (17.689)
23.389/1.9=Voy

 

[collinsmark]

is it 12.31m/s

5.7=Voy (1.9) - .5(9.8)(1.9^2)
5.7=Voy (1.9) - (17.689)
23.389/1.9=Voy

That looks about right (out to the first three significant figures anyway.)

So so now that you know the initial y-component of velcity v_0y, what's the final v_y when it hits the wall? (Back to kinematics.)*

*(At around this point, or a little before, you could use conservation of energy to complete the rest of this problem. But since you've been using kinematics so far, you might as well stick with that, since kinematics works just as well.)

 

[orangem]

Vy^2=Voy^2+2at
Vy^2=12.31^2+2(9.8)(1.9)

Vfy=13.7m/s

 

[collinsmark]

Vy^2=Voy^2+2at
Vy^2=12.31^2+2(9.8)(1.9)

Vfy=13.7m/s

Ummm. :uhh: I think you are confusing two of your kinematics equations, and mixing them [STRIKE]up[/STRIKE] together. (Try a different kinematics formula.)

 

[orangem]

Woops, yeah I did.

vy= v0y - gt
Vy=12.31 -9.8(1.9)

Vfy=6.31m/s

 

[collinsmark]

Woops, yeah I did.

vy= v0y - gt
Vy=12.31 -9.8(1.9)

Vfy=6.31m/s

Okay, that looks correct out of the first three significant figures anyway (ignoring rounding issues on the third significant figure).

(It might be worthwhile at this point though to consider the "sign" of v_fy. You'll end up with the correct final answer anyway, but if you were ever asked to draw the final velocity vector it could be important. Is v_fy moving in the upwards direction or downwards direction? If you're still unsure, try that last step using the kinematics equation v_f = v_i + at, instead of the one that I'm guessing you used with squares in it.)*

Okay, now you have both the final x and y components of the velocity, as the ball hits the wall. You should be able to take it from here and find the magnitude and angle.

*[Edit: if you are wondering where the discrepancy is between the use of the two kinematics equations, remember every time to you take the square root of both sides of an equation, you need to tack on a ± sign to be rigorous. You can't always assume that the result is positive.]

[Another edit: Never-mind, I should have read your previous post more carefully. I see that you already did use the v_f = v_i + at formula. Okay then, so now you should be able to figure out the direction of v_fy, followed by the overall speed and angle. You've finished all the physics. The rest is just geometry. ]