December 11, 2019, 01:01:47 am

### AuthorTopic: Projectile Motion - v^2 = u^2 + 2as  (Read 197 times) Tweet Share

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#### Morrice

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##### Projectile Motion - v^2 = u^2 + 2as
« on: October 28, 2019, 11:07:23 pm »
0
A projectile launched at 100 m/s and at an angle of 30°. Find the magnitude of the velocity at t = 8 seconds.
(attachment 1)

Could someone explain to me why the second method is working?

For v2 = u2 + 2as,
The velocity used is the overall velocity (100m/s), but the acceleration and displacement (ay and sy) are from the vertical component.
Many sources even emphasise the subscripts in the formula (attachment 2).

Oddly enough, this works for all values of time (or any other variable).

Thank you.

« Last Edit: October 28, 2019, 11:11:52 pm by Morrice »

#### DrDusk

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##### Re: Projectile Motion - v^2 = u^2 + 2as
« Reply #1 on: October 28, 2019, 11:41:30 pm »
+4
A projectile launched at 100 m/s and at an angle of 30°. Find the magnitude of the velocity at t = 8 seconds.
(attachment 1)

Could someone explain to me why the second method is working?

For v2 = u2 + 2as,
The velocity used is the overall velocity (100m/s), but the acceleration and displacement (ay and sy) are from the vertical component.
Many sources even emphasise the subscripts in the formula (attachment 2).

Oddly enough, this works for all values of time (or any other variable).

Thank you.
The reason for this is because in Physics we usually say the Horizontal velocity is constant. Which means the only acceleration the projectile experiences is in the y direction. I'll prove it to you
Start off with:
$V^2 = V_x^2 + V_y^2 \\ = U_x ^2 + V_y ^2 \\ \text{but}\hspace{2mm} V_y^2 = U_y^2 + 2a\Delta y\hspace{2mm}\text{Used to Calc vel in the Y direction} \\ \therefore V^2 = \underbrace{U_x ^2 + U_y ^2}_{= U^2} + 2a\Delta y \\ = U^2 + 2a\Delta y$

Since the acceleration in 'x' is always zero the 2a(Delta x) term is completely missing. This means we don't even need to consider the displacement in the x direction!

The formula you gave in the second image is ONLY for the velocity in the Y direction. Whereas the one without the subscripts is for TOTAL velocity, hopefully you can see why it works in my working out. Both formulas are correct but calculate different velocities.

Good question, hopefully this solidifies your understanding =)
« Last Edit: October 28, 2019, 11:54:39 pm by DrDusk »
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#### Morrice

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##### Re: Projectile Motion - v^2 = u^2 + 2as
« Reply #2 on: October 28, 2019, 11:55:52 pm »
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The reason for this is because in Physics we usually say the Horizontal velocity is constant. Which means the only acceleration the projectile experiences is in the y direction. I'll prove it to you
Start off with:
$V^2 = V_x^2 + V_y^2 \\ = U_x ^2 + V_y ^2 \\ \text{but}\hspace{2mm} V_y^2 = U_y^2 + 2a\Delta y\hspace{2mm}\text{Used to Calc vel in the Y direction} \\ \therefore V^2 = \underbrace{U_x ^2 + U_y ^2}_{= U^2} + 2a\Delta y \\ = U^2 + 2a\Delta y$

Since the acceleration in 'x' is always zero the 2a(Delta y) term is completely missing. This means we don't even need to consider the displacement in the x direction!

The formula you gave in the second image is ONLY for the velocity in the Y direction. Whereas the one without the subscripts is for TOTAL velocity, hopefully you can see why it works in my working out. Both formulas are correct but calculate different velocities.

Good question, hopefully this solidifies your understanding =)

Wow!

Yes, that makes perfect sense.
Thank you so much.

-Morrice

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##### Re: Projectile Motion - v^2 = u^2 + 2as
« Reply #3 on: November 17, 2019, 09:57:56 am »
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Hi, could we not just use v = u + at? As we know u, a and t, do we have to break it up into its components? Using this formula, I get a very different answer to what is cited above, but I don't know why it isn't working.

Thanks!

#### fun_jirachi

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##### Re: Projectile Motion - v^2 = u^2 + 2as
« Reply #4 on: November 17, 2019, 10:49:56 am »
+1
Hey there!

By definition, the velocity v of an object in a 2D plane (as considered in projectile motion) is derived from both its vertical and horizontal components. If we just use v=u+at, we're effectively assuming that the vertical acceleration has some impact on the horizontal velocity (which it shouldn't! under this model of projectile motion, there is no effect on horizontal velocity). Your answer doesn't work for this exact reason. Therefore, we must always break it up into components. It's also never this easy and straightforward for any projectile motion question - if it was, something would almost certainly be amiss.

Hope this helps
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