How do you work out the take off velocity from max vertical height attained?

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Discussion Overview

The discussion revolves around calculating the initial velocity required to achieve a specific maximum height when throwing a ball vertically. Participants explore concepts related to energy conservation and kinematics, while also addressing the mathematical expressions involved in these calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents a scenario of throwing a ball to a height of 30 meters and asks how to calculate the initial velocity.
  • Another participant explains the conservation of energy principle, introducing kinetic and potential energy equations, and derives the formula v = √(2gh) for calculating initial velocity.
  • Some participants express difficulty understanding mathematical symbols and formulas, indicating a lack of familiarity with the notation used.
  • A participant clarifies the meaning of the "2" in the formula, explaining it as a multiplier for gravity and height rather than an exponent.
  • Examples are provided to illustrate the application of the formula with specific heights, leading to calculated velocities.

Areas of Agreement / Disagreement

Participants generally agree on the use of the conservation of energy principle and the derived formula for calculating initial velocity. However, there is some uncertainty regarding the understanding of mathematical symbols and the specific application of the formula, indicating a lack of consensus on the clarity of the explanation.

Contextual Notes

Some participants express confusion about mathematical notation and the derivation of the formula, highlighting a potential gap in foundational knowledge that may affect their understanding of the discussion.

Researcher X
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If you throw a ball straight up, say, and it reaches a height of 30 meters before falling straight down and back to your hand, how fast did you throw it up? I understand we can know this from the deceleration that gravity causes.

Could you explain how to calculate these sort of things in laymen's terms, or link to a web calculator for these scenarios?
 
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A very useful basic principle is the conservation of energy in solving problems like these. Basically, two types of energy exist: kinetic energy, associated with movement of objects, and potential energy, associated with position (a ball raised high above your head has the potential to fall down, increasing it's kinetic energy).

The relevant equations look something like this:
KE=\frac{1}{2}mv^2 where m is mass and v is velocity,
PE=mgh Where g is the acceleration due to the Earth's gravity near the Earth's surface (g=9.8m/s^2 in those units), and h is the height above some fixed point.

For a problem like this, I can say that when I initially throw the ball potential energy is zero and all the energy is kinetic. Then, at the peak, I know it's not moving (kinetic energy is zero) and potential energy is some value associated with height. Using the above definitions of the two types of energy:

KE_{initial}+PE_{initial}=KE_{final}+PE_{final} ,which tells us that the total energy is conserved. From the above argument we can say initial potential is zero, and kinetic final is zero, so we get:

\frac{1}{2}mv^2=mgh Which is another way of saying all the kinetic energy is converted to potential.

Simply rearranging the equation and solving for v:
v=\sqrt{2gh}

The key concept is that the change in energy is zero for any two points along the ball's path. Hope that helps :)
 
I don't understand symbols like that, since I haven't done maths since I was 10.

I'm not sure what the formula is yet.
 
Researcher X said:
I don't understand symbols like that, since I haven't done maths since I was 10.

I'm not sure what the formula is yet.

I'm not sure what you're uncomfortable with. Ignoring the derivation, the equation is:
v=\sqrt{2gh}

Where v is the velocity (speed) with which you threw the ball, g is the acceleration due to gravity (9.8 in units of meters per second per second, ~32 in units of feet per second per second), and h is the peak height.

Are you not familiar with the square room symbol?
 
This kinematics equation derives the launch velocity per the max height achieved by the projectile that was launched 100% vertically.

where,

a = acceleration (in m/s^2)
s = distance (in meters)
v = velocity (in m/s, that’s “meters per second” for clarity purposes)
SQR() = square root of value in parenthesis
9.144 meters = 30 feet

v = SQR(2sa)

SQR(2 * 9.144 meters * 9.8 m/s^2) = 13.38739706 m/s

Note: Where gravity is the acceleration, ‘g’ is commonly used in place of ‘a’ (value of ‘g’ is 9.8 m/s^2 for Earth gravity) and ‘h’ (for height) is commonly used in place of ‘s’ (distance), so the equation becomes:

v = SQR(2hg)
 
Nabeshin said:
Are you not familiar with the square room symbol?

I know the square root symbol from a calculator where you can put in 64, press it, and get 8.

Being that It's wrapped around the 2gh, I guess v is the square root of 2gh, but what's the "2" for? I've seen 2 come after things and it means squared, or if it's a 3, it means cubed. I'm not sure why the 2 is before both the gravity and the height.

So, if we say the max height was 10 meters, would the velocity be the square root of 10 and 9.8 plus each other?
 
Researcher X said:
I know the square root symbol from a calculator where you can put in 64, press it, and get 8.

Being that It's wrapped around the 2gh, I guess v is the square root of 2gh, but what's the "2" for? I've seen 2 come after things and it means squared, or if it's a 3, it means cubed. I'm not sure why the 2 is before both the gravity and the height.

So, if we say the max height was 10 meters, would the velocity be the square root of 10 and 9.8 plus each other?

The two is just a number multiplying both the 9.8 and the height. It doesn't really have anything to do with the square root. Here, let me give you an example.

Take your original problem of something that goes 30m up.
v=\sqrt{2*g*h}
So, v=\sqrt{2*9.8*30}=\sqrt{588}\approx 24.5 meters per second

Or, again for the example you just used of 10 meters,
v=\sqrt{2*9.8*10}=\sqrt{196}=14 meters per second

That's basically as much as you need to know for using the formula. Hope this makes sense!
 
Perfect! That makes perfect sense now. Thanks.
 

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