Projectile Motion: velocity and speed

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SUMMARY

The discussion focuses on calculating the speed and velocity of a ball launched from a height H with an initial velocity v₀. The equations derived include the time of flight t = √(2h/g), the horizontal velocity vx = v₀, and the vertical velocity vy = g√(2h/g). The final speed is confirmed to be |v| = √((g√(2h/g))² + v₀²), establishing that speed is indeed the magnitude of velocity. The conversation also emphasizes the importance of using LaTeX for clear mathematical communication.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Familiarity with projectile motion concepts
  • Basic knowledge of LaTeX for formatting equations
  • Grasp of gravitational acceleration (g)
NEXT STEPS
  • Study kinematic equations for projectile motion in detail
  • Learn how to derive equations for velocity and speed in different contexts
  • Explore advanced LaTeX formatting techniques for mathematical expressions
  • Investigate the effects of air resistance on projectile motion
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MathHelpa
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I came across this one problem that asked for speed and velocity and I was wondering what the speed is suppose to be. I read somewhere that its the magnitude to velocity. So I want to see if I am doing this right.

A ball is launched off the top of a building. The initial velocity of the ball is v₀ = (v₀,0) and the initial height is H. What is the velocity and speed at the moment it hits the ground in terms of v₀,H and g?


y = v₀yt + 1/2at2
y = 0 + 1/2at2
t = √(2h/g)

vx = v₀

vy = voy + at
vy = 0 + at
vy = g√(2h/g)

|v| = √((g√(2h/g)2 + v₀2)

θ = tan-1(g√(2h/g/v₀))

v = √(g√(2h/g)2 + v₀2)m/s tan-1(g√(g√(2h/g)/v₀) positive below x axis.

So the question is the speed suppose to be sqrt(g√(2h/g)2 + v₀2)m/s??
 
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Greetings,
Thanks for the headache ;)
Yes, speed is the magnitude of velocity.
Yes the answer is correct though it can be a bit more simplfied...
A hint for any future questions you may want to ask: If you don't know how to use Latex you may want to ensure your equations are at least readable. :D
For denoting power use ^ for denoting multiplication use * and if you are squaring a term with more variables then do something like this (v*w/x)^4
Regards:thumbup:
 
Here's some quick advice on how to use latex for the next time. It will allow you to make your equations clear and legible. Also, you should follow the posting format for next time.

There's two different ways you can execute latex. Inline latex and emphasized latex.

Inline latex looks like this : ##v = v_0t + \frac{1}{2}at^2##

Emphasized latex takes up a whole line and looks like this : $$v = v_0t + \frac{1}{2}at^2$$

To perform inline latex in your posts, simply wrap your text in ## tags like so :

Code:
##v = v_0t + \frac{1}{2}at^2##

To perform emphasized latex, wrap your text in $$ tags like so :

Code:
$$v = v_0t + \frac{1}{2}at^2$$

When writing within the tags, '^' allows you to write superscripts and '_' allows you to write subscripts ( There's s lot more uses for these two symbols, but as time goes on I'm sure you will get used to it ). \frac{numerator}{denominator} allows you to write fractions too.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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