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.

- rock_mit182

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- rock_mit182

|dw:1397053258940:dw|

- rock_mit182

@skullpatrol

- AkashdeepDeb

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

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## More answers

- rock_mit182

\[y=-\frac{ 1 }{ 2 }g t ^{2}+ v _{0} t+ y _{0}\]

- rock_mit182

that one ?

- rock_mit182

but i don't know about the intial velocity

- rock_mit182

i mean, they did not give that value

- AkashdeepDeb

Yes, you have to find it. Hold on for one second. I'll attach one thing.

- rock_mit182

ok

- rock_mit182

\[|v _{x}|= v _{x0} \cos 40°\]
\[|v _{y}|= v _{y0}\sin40°\]

- rock_mit182

\[v =\sqrt{v ^{2} _{x}}+v ^{2}_{y}\]

- rock_mit182

if i use \[\tan 40°\] that would be the initial velocity ?

- AkashdeepDeb

Do you have the answer key?

- rock_mit182

im going to check that . . .

- rock_mit182

it seems like it doesn't have the key answer

- rock_mit182

@johnweldon1993

- AkashdeepDeb

##### 1 Attachment

- rock_mit182

@Mashy

- AkashdeepDeb

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

- rock_mit182

i know them but i have to solve this quickly

- rock_mit182

thanks anyway

- rock_mit182

please anyone tell me how should aproach this problem

- anonymous

so its just me and u huh? :D

- rock_mit182

lol please help! friend

- anonymous

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|

- rock_mit182

i guess in the horizontal component there is not accelaration

- anonymous

correct!

- AkashdeepDeb

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

- rock_mit182

so in certain way i was right about the tan 40°

- rock_mit182

the thing is i don't know where that comes from, i mean the equation

- AkashdeepDeb

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. :)

- anonymous

that comes from the derviation..
just cnosider horizontal and vertical components
and u can do it that way as well!

- AkashdeepDeb

Yes, you were right, not entirely though. tan theta is not the initial velocity but will be used in calculating it. :)

- rock_mit182

is there a way that not need derivation

- anonymous

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

- AkashdeepDeb

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. :)

##### 1 Attachment

- AkashdeepDeb

And there in the equation, y is actually \(\delta y\) giving the change in height. :D

- rock_mit182

|dw:1397056685213:dw|

- rock_mit182

well im going to try in the first way . ..

- anonymous

just do the way i told u.. u ll get it :-/

- rock_mit182

|dw:1397057202246:dw|

- rock_mit182

well in te way you told me i have to multiple by G on the VERTICAL COMPONENT ?

- rock_mit182

@Mashy

- anonymous

i have a hard time understanding ur 'drawings' :P
jsut tell me what answer u get :P

- rock_mit182

lol

- rock_mit182

\[u = \frac{ t }{ -g*\sin 40*3.05 }\]

- rock_mit182

is that the equation for the vertical component ?

- rock_mit182

http://www.twiddla.com/1561200

- rock_mit182

i guess im wrong far so far fromget to the answer

- rock_mit182

\[\frac{ 10\frac{ m }{ s ^{2} }(10m)^{2} }{ (1.05m-10mtan40°)2 \cos40° } = v\]

- rock_mit182

no way i couldn't do it

- rock_mit182

@mashy what did i wrong dude ... xD

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