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

  1. May 10, 2009 #1
    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?
     
  2. jcsd
  3. May 10, 2009 #2

    Nabeshin

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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:
    [tex]KE=\frac{1}{2}mv^2[/tex] where m is mass and v is velocity,
    [tex]PE=mgh[/tex] 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:

    [tex]KE_{initial}+PE_{initial}=KE_{final}+PE_{final}[/tex] ,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:

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

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

    The key concept is that the change in energy is zero for any two points along the ball's path. Hope that helps :)
     
  4. May 10, 2009 #3
    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.
     
  5. May 10, 2009 #4

    Nabeshin

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

    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?
     
  6. May 10, 2009 #5
    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)
     
  7. May 10, 2009 #6
    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?
     
  8. May 11, 2009 #7

    Nabeshin

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    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.
    [tex]v=\sqrt{2*g*h}[/tex]
    So, [tex]v=\sqrt{2*9.8*30}=\sqrt{588}\approx 24.5 [/tex] meters per second

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

    That's basically as much as you need to know for using the formula. Hope this makes sense!
     
  9. May 11, 2009 #8
    Perfect! That makes perfect sense now. Thanks.
     
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