A rock is thrown upward with a velocity of 22 meters per second from the top of a 25 meter high cliff, and it misses the cliff on the way back down. When will the rock be 9 meters from ground level? Round your answer to two decimal places.

The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.

Explanation:

We will use the equations of motion for this.

u = initial velocity of the rock = 22 m/s

g = acceleration due to gravity = -9.8 m/s²

y = vertical position of the rock at a time t = 9 m

y₀ = initial height of the rock = 25 m

t = time it takes for the rock to reach height of 9 m.

(y-y₀) = ut + 0.5gt²

(9 - 25) = 22t + 0.5(-9.8)t²

- 14 = 22t - 4.9t²

4.9t² - 22t - 14 = 0

solving this quadratic equation,

t = 5.055 s or - 0.565 s

Since time cannot be negative,

t = 5.055 s = 5.06 s

Hope this Helps!!!

Explanation:

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pe = weightxheight = 2000x1000=2,000,000joules

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