Question

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.

Answer 1

vo=5.87m/s

Step-by-step 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