Question

Can someone please help me on to determine the initial velocity of a marble that is launched horizontally from a CPO marble launcher?
it tells me that I must also use formulas that are relevant..
Thank you! ☺

Answer 1

@mathstudent55

Answer 2

if you could please walk me through how to do this, because i honestly don't know what to do...

Answer 3

this is what i wrote so far:

Answer 4

This seems like projectile motion, so there are three steps, I think. The first one would be to rearrange the initial velocity equation to have the initial velocity as the subject:
\[v_{0y} = v_{0}\sin \theta_{0}\]
becomes, algebraically,
\[v_{0} = \frac{ v_{0y}}{ \sin \theta_{0} }\]

So now if we know the firing angle and the initial speed in the y direction we can find the initial velocity. But it's unlikely we'll know the initial y velocity without the initial velocity, so there are ways to find that init. y velocity. How about the second kinematic equation?
\[\Delta y = v_{0y}t + \frac{ 1 }{ 2 } g t^{2}\]
We can rearrange this to give us v naught y:
\[v_{0y} = \frac{ \frac{ 1 }{ 2 }g t^{2} - \Delta y }{ t }\]
Now if we have the change in y and time we can get the initial velocity. Does this help? It probably doesn't, but I'm here for questions...

Answer 5

I think it helped me! :) And my assignment says to guess the experimental velocity, so would i just assume that the velocity of the hortizontal path of trajetory would be 0 ?

Answer 6

Not necessarily. The vertical velocity is zero, sorry I didn't realize it was launched horizontally, you'd probably want to change the ys to xes and the sin to cos, the acceleration would be 0 which would eliminate that entire part of the second kinematic equation, and you'd find that:
\[v_{0x} = \frac{ \Delta x }{ t}\]
Which you knew anyway. :) Horizontal firing totally simplifies the whole thing...

Answer 7

oh ok that makes sense! :)

Answer 8

sorry for the low quality but this was the original picture of what was assigned if you would like to read off it...

Answer 9

well thank you so much and I really appreciate your hard in assisting me with this problem and I'm glad i came to use this website and found someone like you :)
Thanks!! I hope have a great day! :)

Answer 10

i meant (hard work)

Answer 11

You're welcome! (Just sayin, if you're new, the best response button is for rewarding the person who helped you the most.)

Answer 12

ok there you go! :)