Question

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

Answer 1

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:

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