Question

Chapter 03, Problem 40 A dlver Springs upward from a board that is 2.80 m above the water. At the instant she contacts the water her speed is 10.9 m/s and her body makes an angle of 82.3 with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction Units (a) Number (b) Number

Answer 1

final speed, u= 10.9 mle Let, initial speed oth - - - - in this line set potential zero. 9999999 ł final energy, Ef = Kinetic energy & potential cany = my + 0 [m - mass of the diver] Initial energy, E = Kinetic energy & potential energy P = £m vo & mgh ; h=2.8m L initial night Due to energy conpervation Eį = Ef + mv. + mgh = mo do + 2gh = Vo = √5 -29h = . J 10.9² - 2x9.8 x 2.8 mip = 7.9956 mp - Vo = 7.9956 mp - magnitude of initial velocity, @ Note, that there is no horizontal force acting on the liver [only gravity actp which in vertical. So, there will be no horizontal component of acceleration, c.e. horizontal component of initial velocity = horizontal component of finel velocity, So, V. Csoo = voso O. = cost (v ceso) i 2 2 & voso Yo- Vocondo 2

$ 0.- con co(02.39] 10.9 7.9956 = 79.47° So, the initial velocity makes an angle 10. = 79.47 | with the horizontal

Please give a thumbs up if you are satisfied with the solution.