Physics

Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, is a baseball, a stopwatch, and a friend on the ground below with a long measuring tape. Bob is a pitcher, and knows that the fastest he can throw the ball is 91.0 mph. Bob starts the stopwatch as he throws the ball (with no way to measure the ball\'s initial trajectory), and watches carefully. The ball rises and then falls, and after 0.910 seconds the ball is once again level with Bob. Bob can\'t see well enough to time when the ball hits the ground. Bob\'s friend then measures that the ball landed 453 ft from the base of the cliff. How high up is Bob, if the ball started from exactly 5 ft above the edge of the cliff?

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asked by Sara
  1. I assume that "level with Bob" means at the same height as it was thrown.
    If he threw at a speed of 91.0 mi/hr = 133.467 ft/s, at an angle θ, then
    the horizontal speed was 133.467 cosθ
    Now, plugging in our distance traveled to landing,
    133.467 cosθ t = 453
    so, it took 453/(133.467 cosθ) seconds to land.
    That means that if the cliff's height is h,
    y(t) = h+5 + 133.467sinθ t - 16t^2
    Now, at time t=0.910, y is again h+5, so
    133.467 sinθ * 0.910 - 16(0.910)^2 = 0
    sinθ = (16*.910)/133.467
    θ = 6.263°
    Thus,
    y(t) = h+5 + 14.56t - 16t^2
    The horizontal speed was 133.467 cos6.263° = 132.670 ft/s
    So, it took 453/132.670 = 3.414 seconds to land.
    So, now we know that
    h+5 + 14.56*3.414 - 16*3.414^2 = 0
    h = 131.778 ft

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    posted by oobleck

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