Aprofessional diver steps off of a cliff that is 18 m high. draw a sketch of the cliff, defining your origin and final position. (careful with negative and positive signs.) unlike the wb assignment, assume the diver jumps up first and has an initial vertical velocity is 4 m/s. (ignore air resistance.) (a) how long does it take the diver to hit the water? (b) what's the diver's velocity on impact with the water? (careful with negative and positive signs.)

19.2 m/s

Explanation:

We set a frame of reference with the origin at the cliff top and the positive X axis pointing down.

Then the initial position is:

X0 = 0

The initial speed is:

V0 = -4 m/s

It is negative because it is speed upwards and the frame of reference is positive downwards.

Since the diver is in free fall, he is affected only by the acceleration of gravity, we can consider him moving under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

He will hit the water at X = 18 m, so:

18 = 0 - 4 * t + 1/2 * 9.81 * t^2

4.9 * t^2 - 4 * t - 18 = 0

Solving this equation electronically:

t = 2.37 s

The diver will hit the water 2.37 s after jumping.

The equation for speed under constant acceleration is:

V(t) = V0 + a * t

V(2.37) = -4 + 9.81 * 2.37 = 19.2 m/s

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