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dlinn [17]
2 years ago
6

A diver springs upward from a board that is 2.70 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 69.2 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.
Physics
1 answer:
TiliK225 [7]2 years ago
7 0

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

vf^{2} =vo^{2}+2*g*h\\

We clear the vo (initial speed)

vo=\sqrt{vf^{2}-2*g*h }

v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}

vo=5.87m/s

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From hook's law,

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