As the elevator at the space needle ascends, it accelerates relative to its position according to...
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As the elevator at the space needle ascends, it accelerates relative to its position according to the following equation a(s) = (-0.085s +3) until it reaches its maximum velocity of 9.75 (s is m m 8 defined from the ground). It stays at the maximum velocity until deccelerating at the top. An engineering student takes a ball up the elevator and releases it out the window at 41m. m a. How fast (in -)is the elevator (and ball) moving when it is released? b. How high (in meters) will the ball go? c. How long will the ball be in the air before it hits the ground? As the elevator at the space needle ascends, it accelerates relative to its position according to the following equation a(s) = (-0.085s +3) until it reaches its maximum velocity of 9.75 (s is m m 8 defined from the ground). It stays at the maximum velocity until deccelerating at the top. An engineering student takes a ball up the elevator and releases it out the window at 41m. m a. How fast (in -)is the elevator (and ball) moving when it is released? b. How high (in meters) will the ball go? c. How long will the ball be in the air before it hits the ground?
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a To find the velocity of the elevator and the ball when the ball is released we need to find the velocity at the position s41 m We can do this by int... View the full answer
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