A diver springs upward from a board that is three meters abovethe water. At the
ID: 1723391 • Letter: A
Question
A diver springs upward from a board that is three meters abovethe water. At the instant she contacts the water her speed is8.90m/s and her body makes an angle of 75.0o withrespect to the horizontal surface of the water. Determine herinitial velocity, both magnitude and direction. answer: 4.52m/s, 59.4o from horizontal A diver springs upward from a board that is three meters abovethe water. At the instant she contacts the water her speed is8.90m/s and her body makes an angle of 75.0o withrespect to the horizontal surface of the water. Determine herinitial velocity, both magnitude and direction. answer: 4.52m/s, 59.4o from horizontalExplanation / Answer
Height h = 3 m Speed at contact v = 8.9 m / s Let the horizontal component of velocity at the time of contact v ' = v cos 75 = 2.303 m / s Vertical component of velocity at the time of contact v" = v sin 75 =8.596 m / s Let the initial velcoity be u and itmakes an angle with horizontal then u cos = v ' = 2.303 ----( 1) In vertical direction : -------------------- Initial velocity U = u sin final velocity V = v " = 8.596 m / s distance S = -3 m accleration a = - g from the relation V ^ 2 - U ^ 2 = 2aS 8.596^2- U ^ 2 = -2g* -3 = 6g = 58.8 U ^ 2 = 15.09 U = 3.884 u sin = 3.884 ---( 2) eq ( 2 ) / eq ( 1 ) ==> u sin / u cos = 3.884 / 2.303 tan =1.6868 = 59.33 degrees fro eq ( 1 ) , initial velocity u = 2.303 / cos59.33 = 4.516 m / sRelated Questions
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