Matthew Mitcham scored 112.10 on his last dive of the 2008 Summer Olympics - a b

Question

Matthew Mitcham scored 112.10 on his last dive of the 2008 Summer Olympics - a back two and a half somersault with two and a half twists - to pull off a stunning upset and win gold.
a. If he started his dive with no vertical velocity, who long did it take him to hit the water?
b. How fast was he traveling vertically when he hit the water?
c. If he hit the water at a point 4 m horizontally out from the takeoff point, what was his horizontal velocity at takeoff and again at the instant he hit the water?

Explanation / Answer

First of all, considering that I couldn't find the hight above the water he starts in this question, lets call the original height h

A) s = ut + 1/2 at2

h = 0 x t + 0.5 x 9.81 x t2

h = 4.905 x t2

t2 = h/4.905

t = (h/4.905)

B) a = (v-u)/t

9.81 = v/((h/4.905))

v = 9.81 x (h/4.905)

C) s = ut

(since gravity only acts in the vertical axis, horizontal velocity isn't affected by any external force, therefore horizontal velocity is constant)

4 = u x (h/4.905)

ux = 4/(h/4.905)

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