Need help solving a tricky projectile motion problem

In summary: But what if youre stuck?In summary, the water spends 3.5 seconds in the air and has a speed of 15cm/s at the point it hits the ground. The initial angle at which the water is shot is undefined.
  • #1
dlacombe13
100
3

Homework Statement


A fountain shoots water from a fountain that is 3.5cm from the ground. The water reaches a maximum of height of 11cm from the ground (or 7.5cm from the height of the fountain). The range of the water is 15cm from the fountain once it hits the ground. Find:
a) The time the water spends in the air
b) The initial speed of the water
c) The initial angle at which the water is shot

Homework Equations


eq.1) Vxi = Vicos(Θ)
eq.2) Vyi = Visin(Θ)
eq.3) Xf = Xi + Vxit
eq.4) Yf = Yi + 0.5(Vyi + Vyf)t
eq.5) Yf = Yi + Vyit - 0.5gt2
*possibly a few more

The Attempt at a Solution


I have been attempting this one for a while now, with no luck. I suspect it involves a system of equations. I have tried first using equations 4 and 5 to solve for t, but I do not think it gave me the right value for t, since I then tried using t in equation 4 to solve for Vyi, but I also do not think that the value of that is right. I tried from there to use more equations, but the angle ended up being undefined or out of the domain of 0<x<90. I am very lost, any tips for at least where to start would be helpful. My professor did leave some hints:
-Solve for the time to its peak, and then for the time back down, and add them.
-Solve for Viy given the information regarding max. height.
But I still can't figure out where to even begin. All I know is that the y-velocity is 0 at the top, Xf=15, Xi=0, Yf=0, Yi=3.5, range or ΔX = 15, and ΔY = 11
 
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  • #2
dlacombe13 said:

Homework Statement


A fountain shoots water from a fountain that is 3.5cm from the ground. The water reaches a maximum of height of 11cm from the ground (or 7.5cm from the height of the fountain). The range of the water is 15cm from the fountain once it hits the ground. Find:
a) The time the water spends in the air
b) The initial speed of the water
c) The initial angle at which the water is shot

Homework Equations


eq.1) Vxi = Vicos(Θ)
eq.2) Vyi = Visin(Θ)
eq.3) Xf = Xi + Vxit
eq.4) Yf = Yi + 0.5(Vyi + Vyf)t
eq.5) Yf = Yi + Vyit - 0.5gt2
*possibly a few more

The Attempt at a Solution


I have been attempting this one for a while now, with no luck. I suspect it involves a system of equations. I have tried first using equations 4 and 5 to solve for t, but I do not think it gave me the right value for t, since I then tried using t in equation 4 to solve for Vyi, but I also do not think that the value of that is right. I tried from there to use more equations, but the angle ended up being undefined or out of the domain of 0<x<90. I am very lost, any tips for at least where to start would be helpful. My professor did leave some hints:
-Solve for the time to its peak, and then for the time back down, and add them.
-Solve for Viy given the information regarding max. height.
But I still can't figure out where to even begin. All I know is that the y-velocity is 0 at the top, Xf=15, Xi=0, Yf=0, Yi=3.5, range or ΔX = 15, and ΔY = 11
You've made a mistake in assigning a value to one of your variables. Yf ≠ 0. How high does the water project from the fountain?

BTW, it's better and more informative to show your actual calculations rather than describe what you did verbally. By examining your calculations, we might be able to determine where you are going wrong.
 
  • #3
This seems fairly simple. For problems involving projectiles, taking the motion along each axis individually helps greatly. Let's look at part (a) and (b), you know delta(y) and the acceleration along that axis ( is it g? ). Using these both can you find Vy and the time taken to reach the top? After that, in the same direction you can find the time taken for the water to reach the ground ( you know the max. height). So this was all in the y direction.
For part (c), you found out Vy in the previous part, so if you could calculate the velocity in the x direction Vx ( is it constant?), you can get the angle. No?
 
  • #4
Either you can find the time easily, by just going to the point that the water is at its maximum fall and treat it as a free falling object. But this is the time only for it going down. How do you think we should get the air time?

Or you can mess around with the equations to get 2 equation solve them and you will end up with the angle then the velocity then you can calculate the airtime (a bit upside down XD)
You will find that sin(theta)2 = sin(2theta)
 
  • #5
Just to add to the info, please remember that the "time of flight" is dictated ONLY by the motion in the vertical direction. In other words, if you find the vertical component of the initial velocity, you will get the same time of flight if you simply shoot the water vertically upwards with that initial velocity and then let it fall to the ground.

That should address part (a) of your question, and you might gain an insight on how to do the rest after that.

Zz.
 
  • #6
Sorry for the delay guys, had to go to work. So I do understand that the time the projectile is in the air only relies on it's y-component of velocity. So for solving part (a), I believe I need to use the equation
Vyf = Vyi - gt ==> 0 = Vyi - (980)t ==> Vyi = 980t
So now I have two unknowns, so I guess I could use
Yf = Yi + Vyit - 0.5gt2 ==> 11 = 3.5 + 980t2 - 490t2
When I solve for t, my calculator gives me about 0.12. Did I do my system of equations correctly? I now need to find the time it takes for it to descend. I saw that someone said Yf ≠ 0. Is it perhaps -3.5 then? I assumed it was zero because I took the ground to be 0, so if the Yi is the height of the fountain (3.5) wouldn't Yf = 0 since the water falls to the ground?
 
  • #7
Why can't you use this kinematical equation? vy2 = vy02 + 2ay

You know the maximum height, and you know that at that height, the y-component of the velocity is zero.

Zz.
 
  • #8
Oh okay, I must have missed that. So I got 121.244 for Vyi. So now I will use that in eq.4 to solve for time. One question I still have is what is the Yf? I know Yi is 3.5cm. So Yi 0 (the ground)?
 
  • #9
dlacombe13 said:
So I got 121.244 for Vyi.
Units?
dlacombe13 said:
One question I still have is what is the Yf? I know Yi is 3.5cm. So Yi 0 (the ground)?
Yes.
 
  • #10
The units are in cm. It seems a little high, but I can't see any error in my calculation. I proceeded to use eq. 5 to find the time:
0 = 3.5 + 121.24t - 0.5(980)t2 = 0.27s
I then used eq. 3 to find Vxi:
15 = 0 + Vxi(.27) = 55.56cm
I then converted the x and y components into the actual velocity vector:
√55.562 + 121.242 = 133.36 cm/s
Finally, I used tan-1(121.24/55.56) = 65.38°
I feel like the speed is kind of high, I don't know why. But do my calculations look correct?
 
  • #11
dlacombe13 said:
The units are in cm. It seems a little high, but I can't see any error in my calculation. I proceeded to use eq. 5 to find the time:
0 = 3.5 + 121.24t - 0.5(980)t2 = 0.27s
I then used eq. 3 to find Vxi:
15 = 0 + Vxi(.27) = 55.56cm
I then converted the x and y components into the actual velocity vector:
√55.562 + 121.242 = 133.36 cm/s
Finally, I used tan-1(121.24/55.56) = 65.38°
I feel like the speed is kind of high, I don't know why. But do my calculations look correct?
Looks about right, but you should not take the output of a calculation step to more sig figs than some of the inputs. Rounding off the time to 0.27s is not a good idea.
You could check the answer by working it backwards: given that speed and angle, should it reach that height and that distance?
 
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Likes dlacombe13
  • #12
Oh yes, you are right. Thank you for the help, along with everyone else who contributed.
 

1. How do I start solving a projectile motion problem?

The first step in solving a projectile motion problem is to identify the given information, including the initial velocity, launch angle, and any other relevant values. Then, use the equations of motion to determine the unknown variables.

2. What are the equations of motion for projectile motion?

The equations of motion for projectile motion are:
- x = x0 + v0cosθ * t
- y = y0 + v0sinθ * t - 0.5gt2
- vx = v0cosθ
- vy = v0sinθ - gt
- v = √(vx2 + vy2)
- θ = atan(vy/vx)
- t = (2v0sinθ)/g
- h = (v02sin2θ)/(2g)
- R = (v02sin2θ)/g

3. How do I handle a problem with an angled launch?

For problems with an angled launch, you will need to break the initial velocity into its x and y components using trigonometry. The x component will remain constant throughout the motion, while the y component will change due to the acceleration of gravity.

4. What is the difference between range and horizontal distance?

Range is the total distance traveled by a projectile, including both the horizontal and vertical components. Horizontal distance, on the other hand, only refers to the distance traveled in the x direction.

5. How do I check if my answer is correct?

You can check your answer by plugging it back into the original equations of motion and making sure it satisfies all the given information. Additionally, you can use a graphing calculator or software to visualize the projectile's motion and see if your answer aligns with the graph.

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