1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the speed (2d)

  1. Jan 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3-46). (a) With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 20m. (b) What is the new minimum speed if the ramp is now titled upward, so that "takeoff angle" is 10 degrees above the horizontal, and nothing else is changed?


    I need help on part b.
    y0 = 1.5m
    x = 20m.
    Z = 10 degrees
    2. Relevant equations
    x = (v0 cos Z) t
    0 = y0 + (v0 sin Z)t - 0.5*g*t2

    3. The attempt at a solution
    I have two equations in two variables, t and v0. So I solved for v0:
    [tex]v_0^2 = (\frac{x^2}{2g\cos^2 Z})(\frac{1}{y_0 + x \tan Z})[/tex]

    So, plugging in, I got 2.05 m/s for v0. But that has to be wrong, especially since my answer in (a) is 36m/s.

    What am I doing wrong?
  2. jcsd
  3. Jan 18, 2007 #2
    I know what I did wrong. I put g in the denominator, but it should be in the numerator....
  4. Jan 18, 2007 #3
    algebra mistake.... the g should be placed on the numerater...
  5. Sep 10, 2008 #4
    How did you develop the equation to solve for the initial velocity?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Find the speed (2d)
  1. 2D Motion Find speed. (Replies: 1)