Question

A diver springs upward from a board that is 4.50 m above the water. At the instant she contacts the water her speed is 14.2 m/s and her body makes an angle of 62.2 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

Answer 1

final horizontal and vertical components of velocities are

V_x=14.2Cos62.2

V_y=-14.2Sin62.2

from kinematic equation

V_y²=V_oy²-2g(y-y_o)

(-14.2Sin62.2)² = V_oy²-2*9.8*(0-4.5)

V_oy=8.34 m/s

V_ox=V_x=14.2Cos62.2=6.623 m/s

A)

Magnitude

V_o=sqrt[V_ox²+V_oy²]=sqrt(6.623²+8.34²)

V_o=10.65 m/s

b)

Direction

o=tan^-1(V_oy/V_ox) =tan^-1(8.34/6.623)

o=51.55^o above the horizontal