Kinematics 2D Projectile Launched at an Angle PROBLEM

[whoopie88]

Homework Statement

A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal, as shown below. Find the ball's hang time, range, and maximum height. Assume air resistance is negligible.

Homework Equations

These are the equations that I have been given by my teacher:
(Final velocity of y/x) = (Initial velocity of y/x) + (Acceleration of y/x)(Time)
(Final velocity of y/x)^2 = (Initial velocity of y/x)^2 + 2(Acceleration of y/x)(Displacement of y/x)
(Displacement of y/x) = (Initial velocity of y/x)(Time) + .5(Acceleration of y/x)(Time^2)

I'm not sure if there are any equations I'm missing, and if I'm missing one, it might be a giant cause of my problem.

The Attempt at a Solution

I have no idea how to solve this equation. I've found the initial vertical and horozontal components of the initial velocity. Vi,x = 23.4 m. Vi,y = 13.5 m. But to find the hang time, range, and height, I'm lost.

Help!

 

[Selim Bradley]

It's easier if you divide the equations into x and y-axis equations:
For x-axis:
Δx = Vix (Δt)
Vfx^2 = Vix^2
For y-axis:
Vf = Vi - g(Δt)
Δy = Viy (Δt) - 0.5g(Δt)^2
Vfy^2 = Viy^2 - 2g(Δy)

I'll help you start the problems, but you should do the calculations on your own:
Your knowns are Vfy = 0 (since at maximum point, velocity equals 0), Viy = 13.5, and a = 9.8 m/s^2 (because in projectile motion, acceleration equals gravity). Which equation should you use to solve for the maximum height? Umm... would you agree that to find the maximum height, Vfy^2 = Vi^2 - 2g(Δy) is the best equation? Since the time is not given, we can't use the other two. Solve for (Δy), you should get the maximum height.

 

[Selim Bradley]

Now that you have Δy, you want to solve for the hang time of the football. Since you have Viy = 13.5, a = 9.8 m/s^2, and Δy, which of the two remaining y-axis equations should you use? Would you agree that Δy = Viy (Δt) - 0.5g(Δt)^2 is the best, since you have enough variables to solve for the hang time? To find the range, or R, you need to find Δx, there is only one equation you can use for that. Use the time that you recently obtained and Vix, and you should be done.

 

[whoopie88]

Thanks for your quick response!

I'm having a lot of trouble with this. I solved for Δy, and I got -37.16. This is the incorrect value for Δy..and I don't see what I did wrong.

Using Vfy^2 = Viy^2 - 2g(Δy), here is the formula with the values substituted in:

0 = 729 + 19.62(Δy)
-729 = 19.62(Δy)
(-729/19.62) = Δy
-37.16 = Δy

But the webassign says it's incorrect. What am I doing wrong?

 

[Selim Bradley]

Viy = 13.5, and that squared (Viy^2) is 182.25, not 729. Thus, when you square a value, if it's negative, negative times negative is always positive. This is how you know you got an error in your procedure.
Also, I forgot to mention that when you use Vfy^2 = Viy^2 - 2g(Δy), you should plug-in gravity as a positive value like this: Vfy^2 = Viy^2 - 2(9.8)(Δy), since the sign was already changed for you (OR you can try using Vfy^2 = Viy^2 + 2a(Δy), and substituting gravity as a negative value, whatever works best for you!). The answer should be 9.30 m for Δy, check me if I'm wrong.

 

[Selim Bradley]

Never mind! I see what you did wrong; remember when using Vfy^2 = Viy^2 - 2g(Δy), you are using Viy (initial velocity in y-plane, OR y-component of Vi), not Vi itself! Disregard what I said about squaring a negative value, you did it right (switching to other side, becomes negative). It's really late, so my head is not working as good right now.

 

[whoopie88]

Oh, I see. I thought the initial velocity in the equation referred to the initial velocity 27 m/s, when it was actually the y component of the initial velocity. I worked it out and solved for Δy and ended up with the same answer as you, 9.29 m.

But there still seems to be a problem..

I solved for t this way, using the equation Δy = Viy (Δt) - 0.5g(Δt)^2:

9.29 = 13.5t - 4.91t^2
0 = -4.91t^2 + 13.5 + 9.29

Using the Quadratic Formula, I got the two solutions: t=1, and 1=2.23. I plugged both of these into the equation for Δx and got 23.4 m and 52.182, neither of which seem to be correct.

What did I do wrong?

(Apologies for asking so many questions, I'm just at a point where I'm just not getting the right answer)

Thank you!

 

[Selim Bradley]

I completely confused you again, sorry! Δy should be zero; if "hang time" is the same as "time of flight." (Since you want to measure the football's entire time on air, it ends up in the ground, therefore Δy = 0). So you should get 0 = -4.91t^2 + 13.5t. You can use distribution to get 0 = t(13.5-4.91t). After this, Δt equals the highest value (13.5) over the smallest (4.91). The reason: it's simple algebra/math . Tell me if dividing those two values gives you the correct answer.

 

[whoopie88]

Oh, thank you. So the value 9.29 for the maximum height is not Δy, but 0 is. Dividing them gave me the right value for Δt, and consequently it gave me the right value for Δx.

Thanks for all your help! I might have to come back later tonight for some more, but in the meantime, I really do appreciate it!

 

[Selim Bradley]

No, the maximum height IS 9.29 m. But Δy will change according to the situation/problem (using Δy as maximum height would give you the time it takes to reach the maximum height or peak). In order to find the total time the football traveled (time of flight), Δy will be 0 (since it hits the ground in the end). I'm glad I was able to help, it can be a little confusing at first, but you'll get the hang of it.
Now I can go to sleep! :zzz: Others will be able to help you if you have further questions.
Good night!

 

[whoopie88]

Thanks for everything! I understand now, I solved another problem with the things you said and I got the right answers. I really appreciate it!

One more thing though, here's another question that is related to the way we solved this one:

A pitched ball is hit by a batter at a 45° angle and just clears the outfield fence, 103 m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance.

This one's a bit more confusing because I feel like I'm only provided with Δx as 103, Vf as 0, and Δy as 0. I don't know what to do with that..Could you give me a quick tip or help on this last question?

 

[Selim Bradley]

Yes! This is a little harder, but never overthink a physics problem.
Everything you need is given to you.
Remember that to find the x-component: Vix = Vicos(45)
And to find the y-component: Viy = Visin(45)

Using the equation for Δx, you can find t in terms of Vi
Like this: Δx = Vix(Δt)
103 = Vicos(45)(Δt)
Δt = 103/(Vicos(45))

Now you can plug in your knowns in this y-equation:
Δy = Viy (Δt) - 0.5g(Δt)^2
0 (because it hit the ground) = Visin(45) x [103/Vicos(45)] - 0.5 x 9.8 x ([103/Vicos(45)])^2
Solve for Vi, using basic algebra. :zzz: Good night!

 

[whoopie88]

I can't solve that. Either I'm doing it wrong or that equation is too difficult for me to solve. I tried multiple times and ended up with several wrong answers.

 

[Selim Bradley]

Next time, show us how you got the wrong answer, like you did on Post #7. If you tell us your procedure, we can help you understand much faster.

0 = Visin(45) x [103/Vicos(45)] - 0.5 x 9.8 x ([103/Vicos(45)])^2
0 = 0.707 Vi x (145.66/Vi) - 4.9 (21218/Vi^2)
The first two Vis cancel and you can multiply 0.707 with 145.66, therefore:
(1,039,682/Vi^2) = 102.98

Can you solve for Vi now?

 

[whoopie88]

Oh, for some reason it didn't occur to me that I could plug in the exact values for sin45 and cos45...so I was having a hard time finding Vi. Yes, I found the correct answer.

Thank you for all your help!