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Calculating Time and Speed From Verticle Height and Acceleration Due to Gravity

  1. Oct 24, 2009 #1
    I have a very simple question. If I know that a projectile reaches a maximum height of 15m from its launch point, how to I work out the original speed and the time of flight given the only force is the downward acceleeration due to gravity. I'm sure its very easy, but I can';t remmeber the equation.
    Many thanks
    Bob
     
  2. jcsd
  3. Oct 24, 2009 #2
    Hi there.

    15m is a relatively small difference in height, so we can simplify the situation by using an approximation of a constant gravitational acceleration [tex]g[/tex].

    First off, let's find the lauch speed. Are you familiar with the equations of motion for constant acceleration? One of these is [tex]2a(x-x_0)=v^2-v_0^2[/tex]. We can use this to find the launch speed. We know that the height difference is 15m meters, and that at 15m, the projectile has zero velocity. Choosing positive y-direction upwards and gravitational acceleration [tex]-9.80\frac{m}{s^2}[/tex],our equation above then becomes

    [tex]2\cdot (-9.80 \frac{m}{s^2})\cdot (15m)=-v_0^2[/tex]

    I'll leave you to compute the value of [tex]v_0[/tex].

    Having found [tex]v_0[/tex], we can use another of the motional equation and a bit of knowledge about motion with constant acceleration to find the flight time. We know that when the projectile reaches it's atarting position again, its velocity will have equal magnitude to it's launch velocity, but opposite sign. We will us the motional equation [tex]v=at+v_0[/tex] to find the flight time. As I explained above, we want to find [tex]t[/tex] when [tex]v=-v_0[/tex]. We plug it into the equation and solve for [tex]t[/tex] and we get

    [tex]t=\frac{-2v_0}{-9.80\frac{m}{s^2}}[/tex]

    Again, I will leave you to compute the value of [tex]t[/tex].

    The important thing is not just to know which equation to apply to a problem, but to fully understand why the equation is applicable and what it means.
     
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