Question

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

Answer 1

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?

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

So they will reach the same height, right?

Answer 6

yes!

Answer 7

(regardless of the V)

Answer 8

yes!

Answer 9

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:

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

Answer 10

(right? )

Answer 11

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

Answer 12

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

Answer 13

So, in general, if we had:

then the horizontal distance is:
\({\rm D}={\bf v}\times \sin(\theta) \times {\bf v} \cos(\theta)\)

Answer 14

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

Answer 15

and Vx is what exactly?

Answer 16

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

Answer 17

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