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bagirrra123 [75]
3 years ago
12

A rock is thrown upward with a velocity of 14 meters per second from the top of a 20 meter high cliff, and it misses the cliff o

n the way back down. When will the rock be 12 meters from the water, below? Round your answer to two decimal places.
Mathematics
1 answer:
daser333 [38]3 years ago
3 0

Answer:

3.35 s after the rock is thrown, it will be 12 m above the water

Step-by-step explanation:

Hi there!

The height of the rock can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rock at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let´s place the origin of the frame of reference on the water so that y0 = 20 m.

Using the equation of height, we can obtain the time at which the rock is at a height of 12 m above the water:

y = y0 + v0 · t + 1/2 · g · t²

12 m = 20 m + 14 m/s · t + 1/2 · (-9.8 m/s²) · t²

subtract 12 to both sides of the equation:

0 = -12 + 20 m + 14 m/s · t - 4.9 m/s² · t²

0 = 8 m + 14 m/s · t - 4.9 m/s² · t²

Let´s solve the quadratic equation using the quadratic formula:

a = -4.9

b = 14

c = 8

t = [-b ± √(b² - 4ac)] / 2a

t = 3.35 s and t = -0.49 s

Since time can´t be negative, the rock will be 12 m above the water 3.35 s   after it is thrown.

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