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evablogger [386]
3 years ago
9

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 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 hit the water?
m/s
Physics
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

a) t= 0.92 s

b) h = 0.46 m

c) v = -6.04 m/s

Explanation:

A)

  • In order to find the total time that the feet are in the air, we must add two times:
  • 1) time needed to reach to the maximum height (t₁)
  • 2) time from when starts to fall from the maximum height until her feet hit the water (t₂)
  • In order to get t₁, we need to take into account that at her highest point, the vertical speed will be zero.
  • Taking for granted the value for the acceleration due to gravity,
  • g = -9.8 m/s2, we can apply the definition of acceleration, and   replacing by the givens, we can find t₁ as follows:

       t_{1} = \frac{v_{o}}{g} = \frac{3.0m/s}{9.8m/s2} = 0.3 s (1)

  • In order to find t₂, we need to find first the highest point above the board, which is indeed what is asked for in b).
  • We can use the following kinematic equation, taking into account that at the highest point, the final velocity vf will be zero.
  • The equation can be written as follows:

        v_{f} ^{2} - v_{o} ^{2} = 2*g*\Delta h  (2)

  • Replacing by the givens, and solving for Δh, we get:

       \Delta h = \frac{v_{o}^{2}}{2*g} = \frac{(3.0m/s)^{2} }{2*9.8m/s2} = 0.46 m (3)

  • The total height over the water will be just the sum of the takeoff point (1.40 m over the water) and the value we found in (3):
  • H = 1.40 m + 0.46 m = 1.86 m
  • Now, we can use the equation that relates the vertical displacement with the time, remembering that v₀=0, as follows:

       H = \frac{1}{2}*g*t^{2}  (4)

  • Replacing by the givens and the value found for H in (4), and solving for t, we get the value of t₂, as follows:

       t_{2} = \sqrt{\frac{2*H}{g} } = \sqrt{\frac{2*1.86m}{9.8m/s2} } = 0.62 s (5)

  • The total time will the sum of t₁ and t₂:
  • t = 0.3 s + 0.62 s = 0.92 s (6)

B)

  • As we have just found, the highest point above the board was at 0.46m above the takeoff point, so the highest point above the board is just 0.46 m.

C)

  • In order to find the velocity when her feet hit the water, we can use the same equation (2), taking into account that v₀=0, and Δh = H= -1.86m.
  • Solving (2) for vf, we get:

        v_{f} = - \sqrt{2*g*H} = -\sqrt{2*9.8m/s2*1.86m} = -6.04 m/s (7)

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