Calculate the displacement and velocity at times of 0.500, 1.00, 1.50, 2.00 and

Question

Calculate the displacement and velocity at times of 0.500, 1.00, 1.50, 2.00 and 2.50 s for rock thrown straight down with an initial velocity of 14.2 m/s from the Verrazano Narrows bridge in New York City. The underside of this bridge is 70.0 m above the water at its center. Assume air resistance is negligible. Choose the positive direction to be pointing down. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 5.00 m/s, and her takeoff point is 1.40 m above the pool. (a) How long are her feet in the air? s (b) What is her highest point above the board? m (c) What is her velocity when her feet hilt the water? m/s

Explanation / Answer

10. initial velocity, u = 5 m/s ( upwards)
   distance from the pool, h = -1.4 m
   acceleration due to gravity, a = 9.8 m/s/s/ downw2ards
   so let the player be in air for time t
   then h = ut - 0.5gt^2
   -1.4 = 5t - 4.9t^2
   4.9t^2 - 5t - 1.4 = 0
   solving for t
a.    t = 1.247 s
b. her highest point above water = 1.4 + h'
   where 2*g*h' = u^2
   h' = 25/2g = 1.274 m
   max height point = h + h' = 2.67 m above the pool
c. when her feet hit the water her velocity is v
   so, v = u - gt
   v = 5 - 9.81*1.247 = -7.233 m/s ( downwards)

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