A community for students.
Here's the question you clicked on:
 0 viewing
idku
 one year ago
I got a question about vector problem. (Please, I just started don't through hard stuff at me).
idku
 one year ago
I got a question about vector problem. (Please, I just started don't through hard stuff at me).

This Question is Closed

idku
 one year ago
Best ResponseYou've already chosen the best response.1Two balls are thrown from the same height as shown. The \(\color{red}{\rm red}\) ball is thrown with velocity \(\color{red}{ \vec v}\) in the vertical direction. The \(\color{blue}{\rm blue}\) ball is thrown with twice the speed of the red ball at an angle of 30\(^\circ\) with respect to the horizontal. Which ball will go higher? dw:1440947270312:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.1The choices they gave me are: A) the red ball B) the blue ball C) the maximum height for both balls is the same D) it depends on the value of v but I really want to figure how to do these.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3hint: the vertical speed of the red ball is V whereas the vertical speed of the blue ball is 2V*sin(30)=2*V/2=V

idku
 one year ago
Best ResponseYou've already chosen the best response.1V \(?\) 2V \(\sin(30)\) V \(?\) 2V \(\sin(30)\) V \(?\) 2V \((1/2)\) V \(=\) V \(\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1So they will reach the same height, right?

idku
 one year ago
Best ResponseYou've already chosen the best response.1I mean V is still greater than zero, just from the problem. Not that we even need to know this. Thank you!! Just one more thing, so for a horizontal motion of some vector \(\bf v\), with a degree of \(\theta\), the ball will go a horizontal distance D of: \({\rm D}={\bf v}\times \sin(\theta)\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3in order to compute the horizontal distance, we have to consider the horizontal component of the speed of the blue ball, whose magnitude is: 2*V cos(30)=sqrt(3)*V

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large {v_x} = 2V\cos 30 = 2V \cdot \frac{{\sqrt 3 }}{2} = \sqrt 3 V\]

idku
 one year ago
Best ResponseYou've already chosen the best response.1So, in general, if we had: dw:1440947970856:dw then the horizontal distance is: \({\rm D}={\bf v}\times \sin(\theta) \times {\bf v} \cos(\theta)\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3no, the horizontal distance is: \[\Large D = {v_x}\Delta t\] where \Delta t is the total flying time

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3Vx is: \[\Large {v_x} = V\cos \theta \]

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large {\rm D_{horizontal }}=\left(\Delta t\right) {\bf v} \cos(\theta)\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3In general, referring to your last drawing, the total flying time is: \[\Large \Delta t = \frac{{2V\sin \theta }}{g}\]

idku
 one year ago
Best ResponseYou've already chosen the best response.1just clarifying for myself

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so teh horizontal distance is: \[\Large D = {v_x}\Delta t = V\cos \theta \cdot \frac{{2V\sin \theta }}{g} = \frac{{{V^2}}}{g}\sin \left( {2\theta } \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3yes! g= 9.8 m/sec^2 or g= 32 feet/sec^2
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.