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rock_mit182
Group Title
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.
 4 months ago
 4 months ago
rock_mit182 Group Title
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.
 4 months ago
 4 months ago

This Question is Closed

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
dw:1397053258940:dw
 4 months ago

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
@skullpatrol
 4 months ago

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

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

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
that one ?
 4 months ago

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

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

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

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

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

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

AkashdeepDeb Group TitleBest ResponseYou've already chosen the best response.0
Do you have the answer key?
 4 months ago

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
im going to check that . . .
 4 months ago

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

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
@johnweldon1993
 4 months ago

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
@Mashy
 4 months ago

AkashdeepDeb Group TitleBest ResponseYou've already chosen the best response.0
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
 4 months ago

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

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
thanks anyway
 4 months ago

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

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

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
lol please help! friend
 4 months ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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
 4 months ago

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

AkashdeepDeb Group TitleBest ResponseYou've already chosen the best response.0
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
 4 months ago

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

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

AkashdeepDeb Group TitleBest ResponseYou've already chosen the best response.0
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. :)
 4 months ago

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

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

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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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
 4 months ago

AkashdeepDeb Group TitleBest ResponseYou've already chosen the best response.0
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 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. :)
 4 months ago

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

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
dw:1397056685213:dw
 4 months ago

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

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

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
dw:1397057202246:dw
 4 months ago

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

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
@Mashy
 4 months ago

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

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

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

rock_mit182 Group TitleBest ResponseYou've already chosen the best response.0
http://www.twiddla.com/1561200
 4 months ago

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

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

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

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