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!

Need help finding final speed of a projectile

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data
    3WL9GiU.png
    Golf ball struck with a velocity of 66 ft/s.
    Part a): determine the speed at which the ball hits point B.
    Part b): determine the time of flight from A to B.

    2. Relevant equations
    For Part b):
    (Vo)x = Vcos(θ)
    (Vo)y = Vsin(θ)
    D = Do + (Vo)x * t
    Y = Yo + (Vo)y * t - (1/2) * g * t^2
    For Part a):
    Vy^2 = (Vo)y^2 - 2 * g * (Y-Yo)
    Vx^2 = (Vo)x^2 - 2 * g * (Y - Yo)
    V^2 = [(V)x^2 + (V)y^2]
    V = [(Vo)x + (Vo)y] / 2

    3. The attempt at a solution
    If there are more effective and less complicated ways of solving the asked questions please let me know.
    I solved for time first (part b).
    -- given an initial velocity Vo in the problem statement of 66 ft/s
    -- from diagram angle between horizontal and initial direction of the ball 10° + 45° = 55°
    (Vo)x = Vcos(θ) -------> (Vo)x = 66cos(55) -----> (Vo)x = 37.89 ft/s
    (Vo)y = Vsin(θ) -------> (Vo)y = 66sin(55) -----> (Vo)y = 54.06 ft/s
    -- solving for distance d
    S = So + (Vo)x * t -------> d * cos(10) = 0 + 37.89 * t ----> d = 38.44 * t

    -- solving for final height of ball (at point B).
    Y = Yo + (Vo)y * t - (1/2) * g * t^2 -----> d * sin(10) = 0 + 54.06 * t - 16.1 * t^2 -----> 6.68 * t = 54.06 * t - 16.1 * t^2
    -- solving for time t
    t = 2.94s This answer was taken as correct in online program.
    -- Plugging in t value in d
    d = 38.44 ft
    -- Plugging in d value in Y = d * sin(10)
    Y = 19.62
    For my attempt at part a) final speed of the projectile at point B I tried two different formulas, but with no success.
    Vy^2 = (Vo)y^2 - 2 * g * (Y-Yo) -----> Vy^2 = 54.06^2 - 2 * 32.2 * (19.62 - 0) -----> Vy = 40.73 ft/s

    Vx^2 = (Vo)x^2 - 2 * g * (S-So) -----> Vx^2 = 37.86^2 - 0 -----> Vx = 37.86 ft/s

    V^2 = [(V)x^2 + (V)y^2]

    V = [(Vo)x + (Vo)y] / 2

    The fourth equation doesn't make much sense to me, and with the correct values inserted answer resulted incorrect. The third gives me a really high number which also resulted incorrect.



     
  2. jcsd
  3. Jan 29, 2016 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For b). Think about the horizontal motion and remember Mr Pythagoras!
     
  4. Jan 29, 2016 #3

    cnh1995

    User Avatar
    Homework Helper

    You could write the equation for vertical displacement from the incline's frame of reference(i.e. angle of projection=θ=45°). For that, you'll need components of gravitational acceleration along and normal to the incline. Equation for vertical displacement will be a quadratic equation in t. From incline's frame of reference, vertical displacement will be 0 when the ball reaches point B. Just solve it to get the time of flight.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Need help finding final speed of a projectile
Loading...