Question

Can someone please help me with this problem? I am BEYOND confused! :/
A 1.8 tall basketball player attempts a goal 11 from the basket that is 3.05 high.
f he shoots the ball at a 47◦ angle, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard? The acceleration of gravity is 9.81 m/s2 .
Answer in units of m/s.
Btw this is a vector problem!

Answer 1

This is what the actual problem looks like- picture included.

Answer 2

well i solved this question today :)
its a 2D motion so we need to find the equation of X and Y
so i got the answer as 11m/s
but always consider a calc mistake so i might be wrong :)

Answer 3

how did you do it?

Answer 4

=..= im to lazy to think noW .. sowwiie >_ <

Answer 5

is it projectial question

Answer 6

yes @uzumakhi

Answer 7

This equation requires 1 kinematic equation, the ones that deal with displacement. Since this is a 2-dimensional problem, I will separate the equations into their x and y components, which will leave a total of 2 equations, which I will label and use to help you solve your question.

1. \[x _{t} = x _{o} + v _{o _{x}}t\]
2. \[y _{t} = y _{o} + v _{o _{y}}t + \frac{ 1 }{ 2 }a _{y}t\]

For the x direction, there is no change in acceleration, it only has initial velocity, so therefore we can ignore the acceleration part of the equation. For the y direction, there is acceleration due to gravity, so we have to keep those there.

Now because this is a 2 dimensional problem, we will have to separate the x and y components for the initial velocity. To do so is simple, just requires a bit of trigonometry. Since the angle is 47 degrees, you can assume that alpha in all future equations will be that.

The y component for the initial velocity is as follows:\[v _{o} \cos \left( \alpha \right) = v_{o_{x}}\]
The x component for the initial velocity is as follows:\[v _{o} \sin \left( \alpha \right) = v_{o_{y}}\]
You can plug these in for the initial velocity for both the x and y equations, then substitute time t with the x equation, by solving for t in equation 1.
\[\frac{ x _{t} }{ v _{o _{x}} } = t\]
Since equation 2 also has a value for t, you can plug it in, which should give you
\[y _{t} = y _{o} + v _{o _{y}}\left( \frac{ x _{t} }{ v _{o _{x}} } \right) + \frac{ 1 }{ 2 }a _{y}\left( \frac{ x _{t} }{ v _{o _{x}} } \right)\]

Now all you have to do is solve for the initial velocity, v sub zero.

Answer 8

Wow that took forever to make hahaha

Answer 9

o_0

Answer 10

The answer that @AskSamphysics provided is roughly correct, just due to calculator error just as he stated, and rounding. But that is the same answer I got as well.

Answer 11

okay! @hkim THANK YOU SO MUCH! You're an angel!

Answer 12

X}

Answer 13

Not a problem @natasha.aries Just give me a Best Response vote and close the question if I answered it to your needs =]

Answer 14

im going to work it out again and make sure i fix my mistake! and then ill close it:)

Answer 15

(\_/)
(O_O)
.(v v)

Answer 16

No problem. Just let me know if you have any other questions while solving this problem.

Answer 17

will do! :)

Answer 18

Oh I just noticed a quick mistake, for the two trigonometric velocity equations, the first one is for the y component, and the second one is for the x component, though the equations themselves have those flipped. Just make sure you do the algebra right, and you should have no problem.

Answer 19

im a little confused about the equations bc when i plug it in it confuses me :/

Answer 20

Which one confuses you? The cosine alpha one is for the y component and the sine alpha one is for the x component

Answer 21

i dont understand how i would solve for it bc i know the degrees but thats it

Answer 22

Basically, you would just replace all the y components of the initial velocity with the trigonometric equation for that component. You are only solving for the following\[v _{o}\] and that is it.

I guess I'll write out another question so you can see it a bit easier.\[y _{t} = \left( v _{o}\sin \left( \alpha \right) \right) t - \frac{ 1 }{ 2 } g t ^{2}\] For all values of t, you just replace it with the following.
\[t=\frac{ x _{t} }{ v _{o}\cos \left( \alpha \right) } \] I'll give you a hint that should help you as well.
\[\frac{ v _{o}\sin \left( \alpha \right) }{ v _{o}\cos \left( \alpha \right) } = \tan \left( \alpha \right) \]

Answer 23

thank youu

Answer 24

Let me know if you still need help solving the problem, this is the last one I'll make sure is solved before I retire for the night =]

Answer 25

lol no go to sleep! if anything ill send it and you can answer it tomorrow!

Answer 26

No worries, this was my plan anyways. I'm learning physics as well on my own, so this is good learning experience for me as well.

Answer 27

are you sure? idk how long it will take! i would much rather you sleep first! and what grade are you actually in then?

Answer 28

I am a college dropout, studying physics for both my own benefit and my future. I wish to be a physics professor someday =P. Therefore, no need to worry about when I'm going to sleep.

Answer 29

i hope your wish comes true! :)

Answer 30

Well I have to take it a step at a time. And the step I'm taking right now is helping you solve this problem, so please do not hesitate to ask if you need further assistance.

Answer 31

Okay!

Answer 32

what is xt in xt/vox=t ?

Answer 33

That is the x position that the ball has traveled in time t. According to the question, that is basically how far the ball traveled along the x-axis, until it reaches the goal.

Answer 34

how do i find that?

Answer 35

It is given to you in the problem. Look closely, you should be able to see it, how far the ball traveled in the x-direction.

Answer 36

11

Answer 37

Correct! How about the y-direction? Make sure to take into account the height of the player as well.

Answer 38

what would i put for vo?

Answer 39

That is what you are finding out for. Vo is the answer to the question, and you will need algebra to get to it.

Answer 40

then what is t?

Answer 41

t is irrelevant. That is the time that it takes for the ball to reach there, but with the given values, is unable to be solved for. Luckily, if you solve one of the first 2 equations for t, then plug that in for t, then you won't ever need to use it.

Answer 42

okay i think im close! whats yo? 0?

Answer 43

Correct! Same for Xo. With the way that we setup the problem, the initial position of both the x and y coordinates of the ball is 0. You can say the ball starts at the origin.

Answer 44

i got 90. 9 :/

Answer 45

Most likely, the algebra was done incorrectly. What values did you get for all the variables?

Answer 46

(.73135)(16.1291)+1/2 (9.81)(16.1291)

Answer 47

Ah, I see where the issue was. You have to realize that since the direction that the Y axis is increasing is up, then that means gravity, would have to be, in fact, negative, for the answer to come up correctly. Try putting a - in front of the g(9.81) to solve for the problem.

Answer 48

i tried that afterwards and got -67.32

Answer 49

I think you might have done the algebra wrong.
\[y=x \tan \left( \alpha \right) + \frac{ 1 }{ 2 }\frac{ g x ^{2} }{\left( v _{o}\cos \left( \alpha \right) \right)^{2}}\] is where you start with. From there, if you do the algebra right, you should get as follows
\[\frac{ 2\left( y-x \tan \left(\alpha \right) \right) }{ g x ^{2} } = \frac{ 1 }{ \left( v _{o} \cos \left( \alpha \right) \right) ^{2}}\] if you take inverse square root, you get
\[v _{o} = \frac{ \sqrt{\frac{ g x ^{2} }{ 2\left( y - x \tan \left( \alpha \right) \right) }} }{ \cos \left( \alpha \right) } \]Just solve for Vo in that equation. Plug in everything as you have it. If you get a negative value, you most likely put the wrong sign on gravity.