A diver springs upward with an initial speed of 2.44 m/s from a 3.0-m board?

Question

A diver springs upward with an initial speed of 2.44 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?

Answer 1

Start with finding the high point - it makes part a simpler.

At the high point speed = 0 m/s, so then

y = (v-v0)^2/(2a) = v0^2/(2g) = 0.303 m from the board.

Distance above the water is 3 m + y = 3.303 m

Now you can use the same logic and equation with v0 = 0, y =3.303 m and a = g ---> solve for v

v = sqrt(2ay) = sqrt(2*9.8*3.303) =8.064 m/s

Answer 2

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Answer 3

♣ since you are just checking your answers, then
v=8.05 m/s; h= 0.304 m;