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Naddik [55]
3 years ago
12

A rock is dropped from the edge of a cliff into a pool of water. Assume free-fall acceleration is 10 m/s per second, and air res

istance can be neglected. If the rock falls for 2.0 seconds before it hits the water, the height of the cliff above the pool is ___________.
Physics
1 answer:
il63 [147K]3 years ago
7 0

Answer:

20 m

Explanation:

From the equation of motion,

S = ut+1/2gt²................................. Equation 1

Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.

Note: Because the rocked is being dropped from a height, acceleration due to gravity is positive (g), and initial velocity (u) is negative

Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)

Substituting into equation 1

S = 0(2) + 1/2(10)(2)²

S = 5(4)

S = 20 m

Hence the height of the the cliff above the pool is 20 m

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The total distance from the graph is calculated as follows;

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The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

  • This shortest distance is the straight line connecting the start and end position. Call this line P
  • From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm
  • Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of  line P.

                                                ↓end

                                                ↓

                                                ↓ 6cm

                                                ↓

  start -------------13 cm------------

Use Pythagoras theorem to solve for P.

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The average velocity of the ant is calculated as;

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Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

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