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idku

  • one year ago

I have another Q - harder than before-:(

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  1. anonymous
    • one year ago
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    what is it?

  2. idku
    • one year ago
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    Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of θ = 35 degrees to the horizontal. The ball is struck 3 feet above home plate. You watch as the ball goes over the outfield wall 420 feet away and lands in the bleachers. After you congratulate Ted on his hit he tells you, "You think that was something, if there was no air resistance I could have hit that ball clear out of the stadium!" |dw:1440953349077:dw| Assuming Ted is correct, what is the maximum height of the stadium at its back wall x = 565 feet from home plate, such that the ball would just pass over it?

  3. idku
    • one year ago
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    Can someone "walk" me step by step?

  4. idku
    • one year ago
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    They also gave me, just in case: 9.8 m/s^2 = 32.2 ft/s^2 1 mile = 5280 ft

  5. anonymous
    • one year ago
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    nice drawing by the way ( the batter) lol

  6. mathmate
    • one year ago
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    Hint: 1. first split the velocity 176 ft/s into horizon and vertical components. vx=176 cos(35) vy=176 sin(35) 2. Ignoring air resistance, find the time t it takes to reach the stadium wall. x(t)=vx*t, where x=565, vx=176cos(34). 3. Find the height of the ball according to the kinematics equation: y(t)=3+vy*t+(1/2)gt^2 (g=acceleration due to gravity = -32.2 ft/s^2, downwards). the 3 ft. at the beginning because the ball is hit at 3 ft above ground. 4. y(t) is maximum height that the ball will clear. (I get over 150 ft.)

  7. Empty
    • one year ago
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    |dw:1440957695997:dw| This is supposed to be a battler fighting a dragon

  8. mathmate
    • one year ago
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    :) nice dragon!

  9. triciaal
    • one year ago
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    |dw:1440957532068:dw|

  10. idku
    • one year ago
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    thanks

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