1. The problem statement, all variables and given/known data A golf ball is hit at an angle of 30 degrees, it travels 50m landing straight in the hole. Ignoring air friction, what is the initial velocity? 2. Relevant equations not too sure, d=vi.t+1/2.a.ts∧2, vf=u∧2+2.a.d, v=u+a.t, d=u.t 3. The attempt at a solution Vertical initial Velocity= usin30 Horizontal initial velocity= ucos30
Could you give some advice in how I would do that? Like which formula to use? Can it be figured out using these formulas?
remember that when there is no air resistance the vertical and horizontal components of the velocity are completely separate from each other, and because there is no air resistance, the horizontal velocity stays the same throughout the flight. P.S. Have I said too much?* *this is to the admins and mentors and other senior members
I tried working it out... Uv= usin30 Uh= ucos30 t= d/Uh, t= 50/ucos30 s=Uv.t+(1/2).-9.8∧2 and does s=0? I can't remember why. 0=usin30.(50/ucos30)+(1/2).-9.8.(50/ucos30)∧2 0=??? how do i simplify this?
maybe try and refer to earlier notes on this, that the book might have talked about, to do with components of vertical/horizontal linear motion…
if s=0 it's simplified to u= (square root of 9.8x50)/cos30 and the answer is 23.78661943 but if s=50 the answer is 27.80110847 What is correct? and why? Please help
okay then. When s=25, then v will be at it's minimum value, and there will be no vertical velocity at all. you can then work out what the vertical distance of s is (with the help of tan 30). from that answer, you can then use t = sv… …then everything else will slowly fall into place.
Vf=Vi+-9.8x(25/Vi.cos30) 0=v-9.8x(25/v.cos30) v=9.8x(25/v.cos30) u∧2cos30=9.8x25 u∧2=(9.8x25)/cos30 u= square root((9.8x25)/cos30) u=16.8196799??
the vertical height can be used to find the flight time for the golf ball half way through the overall flight. With that you can then find the initial velocity of the ball.
If you have the vertical displacement, then it would be this: t=sv As s = 25 m, and at the halfway point of the flight time, the golf ball is at it's highest point, the vertical velocity is 0 ms-^{1}, then you can do the maths. Also, as a reference, here are the formulae.
No, no, it is me. There are other equations: [tex]s=\frac{1}{2}(u+v)t[/tex] [tex]v=u+at[/tex] [tex]s=ut+\frac{1}{2}at^2[/tex] [tex]v^2=u^2+2as[/tex] Which one is the most appropriate? Also, "v" is the final and "u" is the initial velocity. from knowing the answer to the equation you use, you then just need to do a bit of trig… …and there is your answer.
It's important to keep the vertical and horizontal components separate. In the horizontal direction happily you are given the distance X as 50m. The equation "s=Uv.t+(1/2).-9.8∧2" is more clearly written as Y = VoSinθ*t - 1/2*g*t^{2} S or Y as I have rewritten it is 0 because you end at net 0 Y distance up/down. It is important not to confuse X and Y. So in Y, VoSinθ*t = 1/2*g*t^{2} and this yields pretty simply that t = 2VoSinθ/g But you also know from the horizontal X direction that t = X/VoCosθ Now you should find yourself in splendid position to eliminate t as suggested, since you happily have two equations, albeit 1 horizontal and the other vertical, that must have t = t. You don't need to be Tiger Woods to get to the 19th tee now I should hope.