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.

mistakes are ok

-first use of geothermal energy occurred more than 10,000 years ago in north america by american paleo-indians

-the first industrial use of geothermal energy began near pisa, italy in late 18th century

-in 1904, italian scientist piero ginori conti invented the first geothermal electric power plant in which steam was used to generate the power

-with the above experiment, the first geothermal plant in usa started in 1922 with a capacity of 250 kilowatts

-however, in 1946 first ground-source geothermal heat pump installed at commonwealth building in portland, oregon

-during the 1960’s, pacific gas and electric began operation of first large scale geothermal power plant in san francisco, producing 11 megawatts

-in 1973, when oil crisis began many countries began looking for renewable energy sources and by 1980’s geothermal heat pumps (ghp) started gaining popularity in order to reduce heating and cooling costs

-as effect of climate change started showing results, governments of various countries joined hands to fight against it, for which kyoto protocol was signed in japan in 1997, laid out emission targets for rich countries and required that they transfer funds and technology to developing countries, 184 countries have ratified it

hope this