anonymous
  • anonymous
How would you proove this: vx=xsqart(g/2y) using this kinematic equation: 1/2at^2+vt+x in both the x and y directions? (I'll retype it using the equations menu)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I can't seem to get the equation to come out using the equations menu but its one of the kinematic equations for an object in projectile motion, the one with the 1/2at^2 in it.
UnkleRhaukus
  • UnkleRhaukus
\[a=\ddot x =\frac{\text{d}^2x}{\text{d}t^2}=g\] integrate with respect to time \[v=\dot x =\frac{\text{d}x}{\text{d}t}=gt\]again integrate with respect to time \[x={gt^2\over 2}\]
UnkleRhaukus
  • UnkleRhaukus
\[t=\sqrt{\frac{2x}{g}}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

UnkleRhaukus
  • UnkleRhaukus
i havent answered your question
anonymous
  • anonymous
for some reason I can't see your code, it says math processing error. I'll try another internet browser to see if i'm having smilar issues there.
UnkleRhaukus
  • UnkleRhaukus
also note that i assumed acceleration due to gravity as the only force acting on your projectile and that your projectile was initially stationary
UnkleRhaukus
  • UnkleRhaukus
anonymous
  • anonymous
okay that works better but the answer I'm trying to proove is actually: vx(initial velocity)=x[squarert(g/2y)]
UnkleRhaukus
  • UnkleRhaukus
of you mean \[v_x\]
UnkleRhaukus
  • UnkleRhaukus
you wish to prove this\[v_x=x\sqrt{ \small \frac{g}{2y}}\]
anonymous
  • anonymous
I can't see the equation
UnkleRhaukus
  • UnkleRhaukus
makes things difficult
anonymous
  • anonymous
Sometimes when I refresh my page I can see it for a brief moment, then it goes away and shows the "math processing error" message. But from what you have shown me it looks correct.

Looking for something else?

Not the answer you are looking for? Search for more explanations.