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3 years ago

14

A rock is thrown horizontally off a cliff with an initial velocity of 15 m/s. How high was the cliff if the rock lands 22 m from

the base of the cliff?

2 answers:

Helen [10]3 years ago

3 0

Answer:

22 m

Explanation:

Hunter-Best [27]3 years ago

3 0

Answer:

cliff height = 10.55 m

Explanation:

Remarks

This is one of those questions that defies belief. The final vertical velocity is the same as if you just dropped the rock, which is amazing.
Another amazing fact is that the horizontal speed has no acceleration.
And amazing fact number three is the the vertical initial speed is 0.
And finally, the time taken to go horizontally = time to go vertically.

Solution

<u>Time</u>

So the time taken is d = r * t

Remember, this formula can only be used when there is no acceleration.

d = 22 meters

r = 15 m/s

t = ?

t = d / r

t = 22 / 15

t = 1.467

<u>Height of the Cliff</u>

vi = 0 (vertially)

a = 9.8 m/s^2

t = 1.467 seconds The time horizontally and vertically is the same.

d = ?

<u>Formula</u>

d = vi*t + 1/2 a t^2

<u>Solution</u>

d = 0 + 1/2 * 9.8 * 1.467^2

d = 10.55 meters.

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Answer with Explanation:

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