Question

Problem 4.40 - Enhanced - with Feedback On a hot summer day, a young girl swings on a rope above the local swimming hole (Figure 1) When she lets go of the rope her initial velocity is 205 m/s at an angle of 350 above the horizontal You may want to review (Pages 99 - 101) Part A If she is in fight for 0617s, how high above the water was she when she let go of the top VO AE ? h m Figure 1 of 1 Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Provide Feedback 35.00 Casa

Answer 1

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solution velocity 22.05 mis angle 'o'n 35° P ro Vcoso vsino total time = 0.517S time of flight for upper portion - T- 2v sino 2x 2.05% sin 3 5* 20.2345 9.8 so rest time taken to travell hight 'h is f= 0.517 0.234 20 277 s VE r velocity will be equal in magnitude at same level lat point p) whille falling while falling down 2.05 m/s and component of velocity in vertical direction vsino will be responsible for travell hight ', hi Ersino)t + ½ gt2 22.05x sings x 0.277 + x9.8*10.277)" 0.3257 +0.376 So, hight above the water will be So, - = 0.7016 m 0.7016 m CSScanned with CamScanner