## idku one year ago I got a question about vector problem. (Please, I just started don't through hard stuff at me).

1. idku

Two 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|

2. idku

The 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.

3. Michele_Laino

hint: 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

4. idku

V $$?$$ 2V $$\sin(30)$$ V $$?$$ 2V $$\sin(30)$$ V $$?$$ 2V $$(1/2)$$ V $$=$$ V 

5. idku

So they will reach the same height, right?

6. Michele_Laino

yes!

7. idku

(regardless of the V)

8. Michele_Laino

yes!

9. idku

I 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: $${\rm D}={\bf v}\times \sin(\theta)$$

10. idku

(right? )

11. Michele_Laino

in 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

12. Michele_Laino

$\Large {v_x} = 2V\cos 30 = 2V \cdot \frac{{\sqrt 3 }}{2} = \sqrt 3 V$

13. idku

So, in general, if we had: |dw:1440947970856:dw| then the horizontal distance is: $${\rm D}={\bf v}\times \sin(\theta) \times {\bf v} \cos(\theta)$$

14. Michele_Laino

no, the horizontal distance is: $\Large D = {v_x}\Delta t$ where \Delta t is the total flying time

15. idku

and Vx is what exactly?

16. Michele_Laino

Vx is: $\Large {v_x} = V\cos \theta$

17. idku

$$\large {\rm D_{horizontal }}=\left(\Delta t\right) {\bf v} \cos(\theta)$$

18. Michele_Laino

In general, referring to your last drawing, the total flying time is: $\Large \Delta t = \frac{{2V\sin \theta }}{g}$

19. idku

g is -9.8 m/s

20. idku

just clarifying for myself

21. Michele_Laino

so teh horizontal distance is: $\Large D = {v_x}\Delta t = V\cos \theta \cdot \frac{{2V\sin \theta }}{g} = \frac{{{V^2}}}{g}\sin \left( {2\theta } \right)$

22. Michele_Laino

the*

23. idku

yes, that makes sense:)

24. Michele_Laino

yes! g= 9.8 m/sec^2 or g= 32 feet/sec^2

25. Michele_Laino

:)

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