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rock_mit182
 11 months ago
A 2.00mtall 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.
rock_mit182
 11 months ago
A 2.00mtall 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.

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rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0dw:1397053258940:dw

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0You need the equation of a parabola for this. Do you know what the parabola equation for projectile motion is? :)

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0\[y=\frac{ 1 }{ 2 }g t ^{2}+ v _{0} t+ y _{0}\]

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0but i don't know about the intial velocity

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0i mean, they did not give that value

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0Yes, you have to find it. Hold on for one second. I'll attach one thing.

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0\[v _{x}= v _{x0} \cos 40°\] \[v _{y}= v _{y0}\sin40°\]

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0\[v =\sqrt{v ^{2} _{x}}+v ^{2}_{y}\]

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0if i use \[\tan 40°\] that would be the initial velocity ?

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0Do you have the answer key?

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0im going to check that . . .

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0it seems like it doesn't have the key answer

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0I 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

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0i know them but i have to solve this quickly

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0please anyone tell me how should aproach this problem

Mashy
 11 months ago
Best ResponseYou've already chosen the best response.0so its just me and u huh? :D

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0lol please help! friend

Mashy
 11 months ago
Best ResponseYou've already chosen the best response.0so 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

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0i guess in the horizontal component there is not accelaration

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0If 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

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0so in certain way i was right about the tan 40°

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0the thing is i don't know where that comes from, i mean the equation

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0Watch 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. :)

Mashy
 11 months ago
Best ResponseYou've already chosen the best response.0that comes from the derviation.. just cnosider horizontal and vertical components and u can do it that way as well!

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0Yes, you were right, not entirely though. tan theta is not the initial velocity but will be used in calculating it. :)

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0is there a way that not need derivation

Mashy
 11 months ago
Best ResponseYou've already chosen the best response.0YES 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

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0http://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 noncalculus 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. :)

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.0And there in the equation, y is actually \(\delta y\) giving the change in height. :D

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0dw:1397056685213:dw

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0well im going to try in the first way . ..

Mashy
 11 months ago
Best ResponseYou've already chosen the best response.0just do the way i told u.. u ll get it :/

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0dw:1397057202246:dw

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0well in te way you told me i have to multiple by G on the VERTICAL COMPONENT ?

Mashy
 11 months ago
Best ResponseYou've already chosen the best response.0i have a hard time understanding ur 'drawings' :P jsut tell me what answer u get :P

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0\[u = \frac{ t }{ g*\sin 40*3.05 }\]

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0is that the equation for the vertical component ?

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0i guess im wrong far so far fromget to the answer

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0\[\frac{ 10\frac{ m }{ s ^{2} }(10m)^{2} }{ (1.05m10mtan40°)2 \cos40° } = v\]

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0no way i couldn't do it

rock_mit182
 11 months ago
Best ResponseYou've already chosen the best response.0@mashy what did i wrong dude ... xD
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