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idku
 one year ago
Ok, can someone show me ....
idku
 one year ago
Ok, can someone show me ....

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idku
 one year ago
Best ResponseYou've already chosen the best response.1I found: Horizontal component of the velocity: \(\LARGE V_x=11\cos(44)\approx 7.9\) Vertical component of the velocity: \(\LARGE V_x=11\sin(44)\approx 7.6\)  How do I find the maximum height that the bsll goes above the ground?

idku
 one year ago
Best ResponseYou've already chosen the best response.1bsll is supposed to say ball.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4What's the question ?

idku
 one year ago
Best ResponseYou've already chosen the best response.1How do I find the maximum height that the bAll goes above the ground?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you know your equations of motion?

idku
 one year ago
Best ResponseYou've already chosen the best response.1I might as well just say, no I don't....

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4i think we need to know the height of those ppl

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4at least we need to know the height of the person throwing the ball

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is projectile motion correct, so it's better if you label your vector quantities more precisely, so we're dealing with the y  direction here\[\vec a = \frac{ v_{yf}v_{yi} }{ t }\] where \[v_{yf}=0\] is there more information to this, all you gave is a diagram?

idku
 one year ago
Best ResponseYou've already chosen the best response.1f is final, i is initial. just clarifying....

idku
 one year ago
Best ResponseYou've already chosen the best response.1I deleted the question I am so bump  Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

idku
 one year ago
Best ResponseYou've already chosen the best response.1Wait, a is acceleration, but for maximum height, why do we need it? (btw, sorry again and again for deleted the question.)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4The key thing to observe is that the vertical component of velocity will be 0 when the ball reaches the maximum height

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[v_y = u_y  gt\] set that equal to 0 and find the time at which that happens

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(a = g\) because we take upwards as increasing position

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(u_y\) = initial vertical component of velocity

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh. my teacher labelled it \(\LARGE y_{0_{^{_{\huge _x}}}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can label it what ever, long as it makes sense to you and your teacher

idku
 one year ago
Best ResponseYou've already chosen the best response.1or, idk maybe I am confusing labels. I suck at this:)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4thats ridankulous normally we use \(v\) and \(u\) for velocity \(x\) and \(y\) for positions

idku
 one year ago
Best ResponseYou've already chosen the best response.1Yeah it is I that mixed something up (as always)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we never used u, always v_0 xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4v = u + at is how i still remember my kinematics eqns from highschool

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's much easier to

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4it doesn't matter but we all need to agree on same thing..

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE V_y=V_{0y}gt\) \(\LARGE 0=V_{0y}+9.81t\) and what did you say set equal to 0? like this, or am I wrong again?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4looks good and do you know why you're setting that equal to 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4no wait, you flipped the sign

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(\LARGE V_y=V_{0y}gt\) \(\LARGE 0=V_{0y}\color{red}{}9.81t\) and what did you say set equal to 0? like this, or am I wrong again?

idku
 one year ago
Best ResponseYou've already chosen the best response.1(I just want to do this the way my teacher does it, because I got enough confusion already for 100 seasons in advance....)

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh, you said g, so I thought... ok will take a note

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE 0=V_{0y}+9.81t\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1And I am setting it to 0, because V_y is the derivative of D_y (where D is displacement  in ase of y, the height)

idku
 one year ago
Best ResponseYou've already chosen the best response.1So we are already having the derivative of the function that we want to maximize. Right?

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE 0=V_{0y}9.81t\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4remember this : \(g\) is acceleration due to gravity, it equals approximately \(9.81\) on surface on earth, it is always positive.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(a\) is acceleration, in our case \(a = g = 9.81\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large 0=V_{0y}\color{red}{\bf}9.81t\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1so g is pos, g is neg. at least I know that. I am a pes(t)  imist.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(a\) is negative because \(g\) is pointing downward but we are taking "upwards" as increasing position

idku
 one year ago
Best ResponseYou've already chosen the best response.1Ok, and then I just find cthe maximum of our unknown displacement of y, based on the given velocity component (which is the derivative of the displacement)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4right, one thing at a time first find the time at which vertical component of velocity becomes 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4really ? at \(t=0\) the ball surely has nonzero vertical component of velocity : dw:1441290728442:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.1Wait, what is my equation for this? (to answer your question about the tvalue at which the vertical velocity is equivalent to zero.)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Easy, as you said, vertical component of velocity of ball is modeled by the function : \[V_y=V_{0y}9.81t \] you're "given" \(V_{0y} = 11\sin(44)\), therefore the function becomes \[V_y=11\sin(44)9.81t \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4let me finish typing first

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4il let u work it but im not finished yet and i think u need to get some clarifications :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you know that \(V_y\) is \(0\) when the ball reaches maximum height so use that fact to find the "time" at which the ball reaches maximum height

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4once you have the "time" at which ball reaches the maximum height, you can use that to find the required maximum height

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a nice way to work with vertical trajectory you could also still do it the way I was showing above, the main thing is knowing the final velocity of the y component = 0 at maximum height

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4that is all i wanted to say go ahead, set \(V_y\) equal to \(0\) and solve \(t\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE V_y=11\sin(44)9.81t \) \(\LARGE 0=11\sin(44)9.81t \) \(\LARGE 11\sin(44)=9.81t \) \(\LARGE t=\frac{9.81}{11\sin(44)} \)

idku
 one year ago
Best ResponseYou've already chosen the best response.1hope at least that I have managed:)

idku
 one year ago
Best ResponseYou've already chosen the best response.1well that to the power of 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It should be \[t = \frac{ 11\sin(44) }{ 9.81 }\]

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE t=\frac{11\sin(44)}{9.81} \) I am even dumber than what I thought

idku
 one year ago
Best ResponseYou've already chosen the best response.1yes, I am dividing by 9.81 don't know what's up with my Brain

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE t=\frac{11\sin(44)}{9.81} \approx 0.78 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You could actually go ahead and use \[y = v_{0y} t\] now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That will give you your maximum height

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE y=V_{0y}\cdot t =11\sin(44)\cdot 0.78 =5.96\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1calculations did wolfram not me, so that is supposed to be correct if I plugged correctly.

idku
 one year ago
Best ResponseYou've already chosen the best response.1That is the max height....

idku
 one year ago
Best ResponseYou've already chosen the best response.1Let me make an attempt to recap what I have done for maximum height ((My suffering is not equivalent of any sin (of \(\theta\)), I hate this!))  RECAP, MAX HEIGHT \(\LARGE V_y=V_{0y}gt \) \(\LARGE 0=11\sin(44)9.81t \) \(\LARGE t =0.78\) so max height occrs at t=0, and now we plug that into vertical displacement \(\LARGE y=V_{0t}\times t \) \(\LARGE y=11\sin(44)\times 0.78=5.96 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why did you write max height is at t = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't round until the final answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So use the full time rather than 0.78

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE y=\sin^2(44)/9.81\) ?

idku
 one year ago
Best ResponseYou've already chosen the best response.10.78 was really 11sin(44)/9.81 So y is y=11sin(44)×11sin(44)÷9.81

idku
 one year ago
Best ResponseYou've already chosen the best response.1well, if that is right then it is still 5.95

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large y = v_{0y} t  \frac{ 1 }{ 2 } g t^2\) \(\Large y = 11\sin(44)\left(\frac{11\sin(44) }{9.81}\right)  \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2\) http://www.wolframalpha.com/input/?i=11sin%2844%29%C3%97%2811sin%2844%29%2F9.81%29%281%2F2%29%289.81%29%2811sin%2844%29%2F9.81%29%C2%B2 I get 2.97597 > 2.98

idku
 one year ago
Best ResponseYou've already chosen the best response.1And that is not correct either :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hold on, let just use \[h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) }\] we also have to remember the 1.5m forgot about that

idku
 one year ago
Best ResponseYou've already chosen the best response.1I don't think that there is another explanation for an incorrect output from the formula, other than that I failed to find all the values correctly. I don't think formulas are faulty. (I start to despair from thinking that I can think ) HIHPHh9hduhoduhdue89wv9vi

idku
 one year ago
Best ResponseYou've already chosen the best response.1I have to go, I will come back about 2 h later;(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, this should work, the problem was we did not take account of the question, you must read it carefully, and not make assumptions. Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. \[h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) } = \frac{ [11\sin(44)]^2 }{ 2(9.81)} = 2.975...\] but we must add the 1.5m as well to this, so we get 2.975...+1.5 = 4.4759m = 4.5m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you should use this though as we found time for this precisely \[h = \frac{ 1 }{ 2 }g t^2 = \frac{ 1 }{ 2 }(9.81)(0.78)^2 = 2.965....+1.5 = 4.47 = 4.5m\] so it works!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4below equation for vertical position perfectly fine \(\Large y = \color{red}{y_0}+v_{0y} t  \frac{ 1 }{ 2 } g t^2\) \(\Large y = \color{red}{y_0}+11\sin(44)\left(\frac{11\sin(44) }{9.81}\right)  \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When you put your final answer, make sure it's 2 sig digs!

idku
 one year ago
Best ResponseYou've already chosen the best response.1Oh, if it was something on a same level of thinking then I would have probably figured. Yes the 1.5 meters of the ground, because my origin is (0, 1.5) Iambatman tnx! I am checking it and it is.....
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