T = 1.68m/3.62m/s = 0.4641 seconds

Second calculate the vertical height using d = .5at^2

Or H = 0.5gT^2

H = 0.5*9.81*0.4641^2 = 1.056 meters

In that case, if the swimmer leaves the board with a horizontal speed of 3.62 m/s, then that’s his horizontal speed all the way to the water. It’s constant. At constant speed, the simple relation

s = vt

applies. So we can find the time he’s in the air quite easily:

t =- s/v = 1.68 m/3.62 m/s = .464 s

The question then is how far does one fall in .464 s? Well, vertically, the initial speed is 0. All of his initial speed was horizontal, and horizontal doesn’t affect vertical. So, we have

s = (1/2) g t^2 = (1/2) (9.80 m/s^2) (.464 s)^2 = 1.06 m

So that’s how far the board is above the water.

You’re completely right about part (b). It goes back to the particular dead horse I’ve been beating throughout this answer: horizontal speed does not affect vertical speed. Vertical acceleration does not affect horizontal speed.

Martin Gardner would call this an “Aha” experience. It can be confusing at first, but when the light goes on, you’ll wonder why you ever had trouble with it. Just hang in there – it’ll come clear.

Next, there is a formula

s = ut + ½at²

s: distance (to be calculated)

u: intial velocity (vertical velocity in this case, i.e. 0)

a: acceleration (due to gravity)

t: time (from first part)

You can just put in the numbers and it will tell you how far he travelled vertically, so how far up the diving board was.

Vertical initial speed = 0 m/s

So Height = 1/2gt^2 = 1/2*9.8*(0.464)^2 = 1.0549504 m