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A 2.00-m-tall basketball player is standing on the floor 10.0 m from the basket, as in Figure P3.58. If he shoots the ball at a 40.0 angle with the horizontal, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is 3.05 m.

Physics
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|dw:1397053258940:dw|
You need the equation of a parabola for this. Do you know what the parabola equation for projectile motion is? :)

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Other answers:

\[y=-\frac{ 1 }{ 2 }g t ^{2}+ v _{0} t+ y _{0}\]
that one ?
but i don't know about the intial velocity
i mean, they did not give that value
Yes, you have to find it. Hold on for one second. I'll attach one thing.
ok
\[|v _{x}|= v _{x0} \cos 40°\] \[|v _{y}|= v _{y0}\sin40°\]
\[v =\sqrt{v ^{2} _{x}}+v ^{2}_{y}\]
if i use \[\tan 40°\] that would be the initial velocity ?
Do you have the answer key?
im going to check that . . .
it seems like it doesn't have the key answer
I can explain you the whole thing if you want. But this video, might just clear ALL the queries you have on this topic. Hold on. https://www.youtube.com/watch?v=wDyfD0EAKU8 https://www.youtube.com/watch?v=kc2eCgHtWK0 https://www.youtube.com/watch?v=5WxdXmNrQdc https://www.youtube.com/watch?v=DQRsK_AaSK4 These are the links from an MIT course called 8.01 that I had taken last year. The videos are fantastic! :D
i know them but i have to solve this quickly
thanks anyway
please anyone tell me how should aproach this problem
so its just me and u huh? :D
lol please help! friend
so the basket is vertically 1.05 m high, and 10m away.. and the angle is 40 thats all that matters to us.. we can forget about everything else now |dw:1397055972696:dw|
i guess in the horizontal component there is not accelaration
correct!
If you have to solve it quickly. Just use the formula then. y = y_0 + x. tan \(\theta\) - \(\frac{g.x^2}{2.v.cos \theta}\) y = height at landing y_0 = initial height x = distance between both heights This is basically the equation of the parabola in projectile motion. http://en.wikipedia.org/wiki/Trajectory_of_a_projectile
so in certain way i was right about the tan 40°
the thing is i don't know where that comes from, i mean the equation
Watch the first video. The equation derivation is very simple. It uses the simple equations of motion. But it considers it in the x and y axis both. :)
that comes from the derviation.. just cnosider horizontal and vertical components and u can do it that way as well!
Yes, you were right, not entirely though. tan theta is not the initial velocity but will be used in calculating it. :)
is there a way that not need derivation
YES YES.. just decompose buddy lets do this let inital velocity be u then in horizontal we have intial velocity = u cos(40) distance = 10m so time = ? u can calculate lets call that as t now u know in the vertical intial velocity = usin(40) acceleration = -g (upward positive) displacement = +1.05 time = t get an expression .. and solve for u xD tadaaaaa
http://www.triton.edu/uploadedFiles/Content/Resources_and_Services/Services/Academic_Success_Center/About_Us/projectiles.pdf You do not need derivation for this, if you are allowed to use the equation the directly. The non-calculus proof is linked above ^ . I can tell you WHY we need the equation. As we have an incomplete parabola, due to the height difference, a @Mashy has drawn above, it is necessary to use the parabola equation because the velocity then, can be calculated at that point. :) This is the equation. This equation is the equation used to get all the other equations of height and range. :)
And there in the equation, y is actually \(\delta y\) giving the change in height. :D
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well im going to try in the first way . ..
just do the way i told u.. u ll get it :-/
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well in te way you told me i have to multiple by G on the VERTICAL COMPONENT ?
i have a hard time understanding ur 'drawings' :P jsut tell me what answer u get :P
lol
\[u = \frac{ t }{ -g*\sin 40*3.05 }\]
is that the equation for the vertical component ?
http://www.twiddla.com/1561200
i guess im wrong far so far fromget to the answer
\[\frac{ 10\frac{ m }{ s ^{2} }(10m)^{2} }{ (1.05m-10mtan40°)2 \cos40° } = v\]
no way i couldn't do it
@mashy what did i wrong dude ... xD

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