Determine her initial velocity, both magnitude and direction.?

Question

A driver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.40 m/s and her body makes an angle of 83.0 degrees with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.

magnitude_________m/s

direction__________degrees

Answer 1

Let:

u be the take off speed at angle a to the horizontal,

v be the final speed at angle b to the horizontal,

m be the mass of the diver,

h be the height difference between the diving board and the water,

g be the acceleration due to gravity.

Horizontally:

u cos(a) = v cos(b) ...(1)

Equating initial PE + KE to final KE:

mu^2 / 2 + mgh = mv^2 / 2

u^2 = v^2 - 2gh

u = sqrt(9.40^2 - 2 * 9.81 * 3)

= 5.43 m/s.

From (1):

5.43 cos(a) = 9.40 cos(83)

cos(a) = 9.40 cos(83) / 5.43

a = 77.8 deg.