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anonymous

  • one year ago

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 about 33.3 m/s. 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 0.710 seconds the ball is once again level with Bob. Bob can\'t see well enough to time when the ball hits the ground. Bob

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  1. anonymous
    • one year ago
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    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 about 33.3 m/s. 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 0.710 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 125 m from the base of the cliff. How high up is Bob, if the ball started from exactly 2 m above the edge of the cliff?

  2. anonymous
    • one year ago
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    Acceleration due to gravity when it goes up is -9.8m/s^2 \[v _{f}=v _{i}+at\] Vf is final velocity and that would be zero. |dw:1443585665673:dw| and since throwing it back up and falling at the same distance(Bob's level) is the same 0.710s=2t t=0.710s/2 Plug that into the first equation I showed you to find initial velocity.

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