Your friend throws a rock off of a cliff straight down with an initial speed of 7.50 m/s
. Using a stopwatch, you determine it takes the rock 4.50 s
to hit the ground below. How high is the cliff?

To find the height of the cliff, we can use the kinematic equation that relates displacement, initial velocity, acceleration, and time:

\[
d = v_i t + \frac{1}{2} a t^2
\]

Where:
- \( d \) is the displacement (the height of the cliff),
- \( v_i \) is the initial velocity (\( 7.50 \, \text{m/s} \)),
- \( a \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)),
- \( t \) is the time it takes for the rock to hit the ground (\( 4.50 \, \text{s} \)).

Now, substituting the given values into the equation:

\[
d = (7.50 \, \text{m/s}) (4.50 \, \text{s}) + \frac{1}{2} (9.81 \, \text{m/s}^2) (4.50 \, \text{s})^2
\]

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Your friend throws a rock off of a cliff straight down with an initial speed of 7.50 m/s. Using a stopwatch, you determine it takes the rock 4.50 s to hit the ground below. How high is the cliff?

Learn how to calculate the height of a cliff using the kinematic equation for motion under gravity. This problem involves an initial downward velocity of 7.50 m/s and a time of 4.50 seconds to hit the ground. Suitable for high school physics students.

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