A cliff diver positions herself on a cliff that angles downwards towards the edge. The length of the top of the cliff is 50.0 m and the angle of the cliff is θ = 21.0° below the horizontal. The cliff diver runs towards the edge of the cliff with a constant speed, and reaches the edge of the cliff in a time of 6.10 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 30.0 m before hitting the water.

After leaving the edge of the cliff how much time does the diver take to get to the water?

How far horizontally does the diver travel from the cliff face before hitting the water?

Remember that the angle is at a downward slope to the right.

After leaving the edge, it takes the diver 2.79 s to reach the water.

She travels 21.4 m horizontally from the cliff.

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

The equation of the position vector of the diver at time t is the following:

r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

θ = angle of falling.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s²).

If we place the origin of the frame of reference at the top of the cliff on the edge, then x0 and y0 are equal to zero (initial position r0 = (0, 0)).

First, let's calculate the initial velocity of the diver knowing that she traveled 50.0 m in 6.10 s at constant speed:

v = d/t

where:

v = speed.

x = traveled distance.

t = time.

v = 50.0 m/ 6.10 s

v = 8.20 m/s

The initial velocity is 8.20 m/s.

We know that the vertical component of the position vector at the time when the diver hits the water is -30.0 m (see r1y in the figure). Then, using the equation of vertical position we can calculate the time it takes the diver to reach the water:

y = y0 + v0 · t · sin θ + 1/2 · g · t²    (y0 = 0)

-30.0 m = 8.20 m/s · sin (21.0°) · t - 1/2 · 9.8 m/s² · t²

0 = 30.0 m + 8.20 m/s · sin (21.0°) · t - 4.9 m/s² · t²

Solving the quadratic equation:

t = 2.79 s (the other solution is rejected because it is negative).

After leaving the edge, it takes the diver 2.79 s to reach the water.

Now, let's use the equation of the horizontal position and find the horizontal distance traveled in 2.79 s:

x = x0 + v0 · t · cos θ    (x0 = 0)

x = 8.20 m/s · 2.79 s · cos (21.0°)

x = 21.4 m

She travels 21.4 m horizontally from the cliff.

trig 15.3 cos

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