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Kinematics Problem

  1. Jul 10, 2015 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/BPPbwpu

    2. Relevant equations
    s = ((v + u)*t)/2 - an equation of motion for constant acceleration, which is applicable in this situation.

    3. The attempt at a solution
    Let u be the velocity at which the projectile was launched.

    It is given that sx = symax. So:

    sx = symax = u*cos(θ)*t

    Thus:

    symax/t = u*cos(θ)

    Now, in the y-direction, the projectile was given initial velocity u*sin(θ), and at the top of its trajectory (where it has maximum height symax), it has final velocity 0 m/s. Thus:

    symax = ((0 + u*sin(θ))*t)/2

    , which means:

    symax/t = u*sin(θ)/2

    However, since symax/t = u*cos(θ) :

    u*cos(θ) = u*sin(θ)/2

    Thus:

    cos(θ) = sin(θ)/2

    2*cos(θ) = sin(θ)

    2 = sin(θ)/cos(θ)

    2 = tan(θ)

    θ = tan-1(2) = 63.4349°

    However, this was marked wrong.

    Can somebody see what could have gone wrong?

    Thank you.
     
  2. jcsd
  3. Jul 10, 2015 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The time to reach the highest point is not equal to the whole time of flight.
     
  4. Jul 10, 2015 #3
    Ahh... I got it. It takes only half the total travel time to reach the top of the trajectory. Thank you.
     
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