Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Projectile motion problem

  1. Nov 18, 2005 #1
    Hey!

    I always had a few difficulties with projectile motion problems, so I just solved one and I wanted to verify if my solution is ok, since I don't have a book with solutions so I can't check if I understood the problem well....

    So it's about somebody on a bike who rides off an entrenchment (that's what it's called right?) with a velocity v under an angle of alpha with the ground. He's hoping to land safely on another entrenchement that's h heigher than the first one, at a distance x from the first entrenchment:

    Projectilemotion.GIF

    For a given height h, find the minimal velocity vmin the jumper needs to have in order to land safely on the platform at a distance x.

    Well what I did is the following:

    The well known formula's for the projectile motion are:
    x(t)= v0x t + x0
    y(t)= y0 + v0y t - 1/2 gt^2
    Where v0x= v0 cos @ and v0y= v0 sin @

    So if x(t)= x than y(t)= h.

    x= v0x t and h= v0y t - 1/2 gt^2
    Therefore t= x/v0x. Substitution in the h formula gives:

    h= ((v0y x)/v0x)- 1/2 g (x/v0x)^2= ((v0y x)/v0x)- (g x^2)/(2 v0x^2)

    Knowing v0x= v0 cos @ and v0y= v0 sin @ substitution gives:

    h= ((x v0 sin@)/ v0 cos @) - (g x^2)/ (2 v0^2 (cos^2)@)
    h= x tan @ - (g x^2)/ (2 v0^2 (cos^2)@)

    (g x^2)/ (2 v0^2 (cos^2)@)= x tan@ -h
    (2 v0^2 (cos^2)@)= (g x^2)/ (x tan@-h)
    v0^2= (g x^2)/ (2 (cos^2)@ (x tan@ -h))
    v0= sqrt((g x^2)/ (2 (cos^2)@ (x tan@ -h)))

    Is this correct?! Thanks in advance for your effort!
     
  2. jcsd
  3. Nov 18, 2005 #2
    Last edited by a moderator: May 2, 2017
  4. Nov 18, 2005 #3

    daniel_i_l

    User Avatar
    Gold Member

    It looks like you understood the problem fine. Your answer also looks good but I didn't have time to go through all the details.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook