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:
[tex]t_{1} = \frac{v_{o}}{g} = \frac{3.0m/s}{9.8m/s2} = 0.3 s (1)[/tex]
- 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:
[tex]v_{f} ^{2} - v_{o} ^{2} = 2*g*\Delta h (2)[/tex]
- Replacing by the givens, and solving for Δh, we get:
[tex]\Delta h = \frac{v_{o}^{2}}{2*g} = \frac{(3.0m/s)^{2} }{2*9.8m/s2} = 0.46 m (3)[/tex]
- 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:
[tex]H = \frac{1}{2}*g*t^{2} (4)[/tex]
- Replacing by the givens and the value found for H in (4), and solving for t, we get the value of t₂, as follows:
[tex]t_{2} = \sqrt{\frac{2*H}{g} } = \sqrt{\frac{2*1.86m}{9.8m/s2} } = 0.62 s (5)[/tex]
- 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:
[tex]v_{f} = - \sqrt{2*g*H} = -\sqrt{2*9.8m/s2*1.86m} = -6.04 m/s (7)[/tex]
