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A shotputter throws the shot with an initial speed of 15.5 m/s at a 32.0 degree angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 m above the ground.

  • physics -

    Split the initial velocity (u) into two components, vertical (uy) and horizontal (ux) using:
    ux=u cos(θ) and
    uy=u sin(θ)

    Solving the time (t) the shot stays in the air from:
    Sy=uy*t - (1/2)gt²
    where Sy = vertical distance travelled = -2.2m
    (reject the negative root).

    The horizontal distance travelled is then
    Sx=ux*t

  • physics -

    What do you mean by where Sy = vertical distance traveled = -2.2m? Is the -2.2m from the equation above? Do I add that in to the equation?

  • physics -

    -2.2m is the distance between the hand and the ground.
    So we equate the vertical distance (from the hand) with the distance as a function of time:
    -2.2=uy*t - (1/2)gt²
    Solve for t.

  • physics -

    24.1 m

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