Question

A diver springs upward from a board that is 4.10 m above the water. At the instant she contacts the water her speed is 12.7 m/s and her body makes an angle of 78.8 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

Answer 1

Let horixontal be x axis and vertical be y axis.

Vf = 12.7 m/s @ 78.8 degree with horizontal

x component of Vf = -12.7 cos 78.8 = -2.467 m/s

y component of Vf = -12.7 sin 78.8 = -12.458 m/s

horizontal component willnot change through out the motion.

x component of Vi = = -2.467 m/s

Now to caculate vertical component of her initial velocity

Vf^2 = vi^2 + 2*g*h

12.458^2 = Vi^2 + 2*9.8*4.1

Vi = 8.65 m/s

So,y compoenent of initial velocity = 8.65 m/s

x component of Vi = = -2.467 m/s

magnitude = sqrt (2.467^2 + 8.65^2)

= 9 m/s

Angle from horizontal = atan (8.65/2.367)

=75 degree