Recent content by 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. :biggrin: 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...

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)...

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...

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 =...

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). :blushing...

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...

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...

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...

I am coming to that conclusion :approve:. However, before telling my professor, it would be nice if someone in the Physics Forums could check my work and OK it first, if that's alright. o:)

Well... I exactly followed their whole procedure. I'm getting the wrong answer still, the correct answer should be 2570 m/s. Thank you for the link though! Anyone else? Help :cry:

Are you sure? That's similar to what I did, and that's not the answer, as I stated above.

Hello everyone! Please, help me understand the following problem: Homework Statement Earth's orbit around the Sun is nearly circular. The period is 1 yr = 365.25 d. (a) In an elapsed time of 5 d, what is Earth's angular displacement? Solved! (b) What is the change in Earth's velocity, Δv...