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 11.7 m/s and her body makes an angle of 64.4 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

Answer 1

(a) Final velocity of the diver, V = 11.7 m/s

Angle with respect to horizontal, = 64.4 deg

Now,

The vertical velocity, when the diver hits the water = 11.7*sin64.4 = 10.55 m/s.

The horizontal velocity = 11.7*cos64.4 = 5.06 m/s

The vertical velocity says that the maximum height she/he had reached was

h = Vy^2/(2g) = 10.55^2/(2*9.81) = 5.67 m

which is (5.67 - 2.70) = 2.97 m above the board.

using again h = Voy^2/(2g)

=> Voy = √(2h*g) ( with Voy = initial vertical velocity)

=> Voy = √(2*2.97*9.81)

=> Voy = 7.63 m/s, which was her initial vertical velocity

Horizontal velocity remains constant.

So, initial horizontal velocity, Vox = 5.06 m/s

Therefore,

Initial velocity of the diver, Vo = sqrt[Vox^2 + Voy^2]

= sqrt[5.06^2 + 7.63^2]

= 9.15 m/s (Answer)

(b) Direction of the velocity, = tan^-1(Voy / Vox)

= tan^-1(7.63/5.06) = 56.45° with the horizontal (Answer)