1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Projectile Motion w/ unknown initial velocity & uneven launch/landing points

  1. Jun 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A baseball thrown at an angle of 60.0 above the horizontal strikes a building 18.0m away at a point 6.00m above the point from which it is thrown. Ignore air resistance.

    A) Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown).
    B) Find the magnitude of the velocity of the baseball just before it strikes the building.
    C) Find the direction of the velocity of the baseball just before it strikes the building.

    initial angle = 60o
    x = 18, y = 6 @ point of impact
    a = 9.8 m/s2

    2. Relevant equations

    x = v0x*t
    y = v0y*t + .5*a*t2

    3. The attempt at a solution
    I began by manipulating the equations above to create an equation that solves for the magnitude of the velocity, v at x=18, y=6 :

    18 = cos(60)*v*t
    6 = sin(60)*v*t + .5*a*t2

    t = 18/(cos(60)*v)
    6 = sin(60)*v*(18/(cos(60)*v))+.5*-9.8*(18/(cos(60)*v))2
    v = sqrt((.5*-9.8*182)/(6-(18*sin(60)/cos(60))))/cos(60)

    And using this I found v to be 15.9, which is incorrect. I don't think the rest will be any problem if I can figure out v, but I can't find what I'm doing wrong for the life of me. Also, if there is a different approach I should be taking please tell me, thanks!
  2. jcsd
  3. Jun 14, 2009 #2
    Try using the position kinematic equation y=-1/2gt^2+vsin(60)t+yo and set the final value (y)at six meters. And the equation x=-1/2gt^2+vcos(60)t+xo, since there is no acceleration in the x vector the equation becomes 18=vcos(60)t, therefore v=18/cos(60)t sub it into the y position equation. You get 6=-4.9t^2+(18*sin(60))/(cos(60)), or t=sqrt((18tan(60)-6)/4.9). Then sub the time back into the v=18/cos(60)t equation. V=9*sqrt((18tan(60)-6)/4.9)
    Last edited: Jun 14, 2009
  4. Jun 14, 2009 #3
    Doesn't this lead to a quadratic equation with t? That's what I ended up with and it's getting messy and even harder to follow...unless I'm misunderstanding you?
  5. Jun 14, 2009 #4
    Sorry about the first post I edited it, you were on the right track just use substitution
  6. Jun 14, 2009 #5
    Ah, thank you...I just realized that I was entering 15.9 into the wrong blank on the submission form...I'm resisting the urge to slam my fist through my wall.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook