What Are the Next Steps in Solving This Physics Problem?

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The discussion revolves around solving a physics problem involving vertical and horizontal motion. The vertical calculations yield a final velocity of 15.94 m/s and a time of 1.626 seconds, while the horizontal motion results in a velocity of 4.613 m/s. For the box being pushed, the time calculated is 1.083 seconds. The key equations used include v^2 = vo^2 + 2aDelta x and Delta x = vxt. The conversation emphasizes breaking down the problem into components and applying the correct formulas to find the unknowns.
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Ok...the puzzler. I haven't gotten really far. But this is what I have...

Vertical:
Delta X= -12.96 m
Vo= 0m/s
a= -9.80 m/s-squared-

v-squared- = vo-squared- + 2aDelta x
v= 15.94 m/s

v = vo + at
t= 1.626 s

Horizontal:
Delta x= 7.5 m
t= 1.626 s
v= ?

Delta x = vxt
vx= 4.613 m/s

Box being pushed:
V= 4.163 m/s
Delta x= 5.0 m
t=?

Delta x= vxt
t= 1.083 s

What do I do from here? I really need help...as I have stated before. Could you please just post the whole process and then I can see if I can get it.

-Paige and fellow Physics 30 Canadian students
 
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To solve for the remaining variables, we'll need to use the equation: v = vo + at. Plugging in the values for vo, a and t, we can solve for the final velocity (v): v = 0 + (-9.8 m/s2) x (1.626 s) = -15.94 m/s For the horizontal motion, we can use the same equation and solve for the initial velocity (vo): vx = 7.5 m / 1.626 s = 4.613 m/s vo = 4.613 m/s - (-9.8 m/s2) x (1.626 s) = 20.13 m/s Finally, for the box being pushed, we can use the equation Delta x = vxt to solve for the time (t): t = 5.0 m / 4.163 m/s = 1.083 s
 


Hi Paige and fellow Physics 30 Canadian students,

It looks like you have made some good progress so far! Let's break down the problem and see if we can figure out the next steps.

First, we have the vertical component of the motion, where we are given the change in position (Delta X), the initial velocity (Vo), and the acceleration (a). Using the formula v^2 = vo^2 + 2aDelta x, you were able to solve for the final velocity (v) to be 15.94 m/s. Great job!

Next, we have the horizontal component, where we are given the change in position (Delta x) and the time (t). We need to find the velocity (v) in this case. Using the formula Delta x = vxt, we can rearrange it to solve for v: v = Delta x/t. Plugging in the values given, we get v = 7.5m/1.626s = 4.613 m/s. Again, well done!

For the final part, we are looking at the motion of the box being pushed. We are given the velocity (v) and the change in position (Delta x), and we need to find the time (t). Using the same formula as before, Delta x = vxt, we can rearrange it to solve for t: t = Delta x/v. Plugging in the values given, we get t = 5.0m/4.163 m/s = 1.083 s.

So to summarize, the process for solving these types of problems is to first identify the given values for each component (vertical and horizontal), use the appropriate formula to solve for the unknown variable, and then plug in the values to get your final answer.

I hope this helps and good luck with your Physics 30 studies!
 
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