Question

A soccer ball is kicked horizontally off the side of a cliff with an initial velocity of 24.0 m/s. If it lands 58.0 m from the base of the cliff, how high is the cliff? (note: the answer is positive since it's asking for height)

Answer 1

Given:

Initial velocity = 24.0 m/s

Distance fron where the ball lands to the base of the cliff = 58.0 m

Let's find the height of the cliff.

Let's make a sketch representing this situation:

Let's first find the time the ball used to go from the top of the cliff to the ground:

`\begin{gathered} t=(d)/(v) \\ \\ t=\frac{58.0\text{ m}}{24.0\text{ m/s}} \\ \\ t=2.417\text{ seconds} \end{gathered}`

Now, apply the kinematics equation to find the height of the cliff, dy:

`\Delta dy=v_it+0.5at^2`

Where:

vi is the initial velocity in the vertical axis = 0 m/s

a is the acceleration due to gravity = -9.8 m/s² (downwards)

t is the time = 2.417 seconds

Thus, we have:

`\begin{gathered} d_y=0(2.417)+0.5(9.8)(2.417)^2 \\ \\ d_y=28.62\text{ m} \end{gathered}`

Therefore, the height of the cliff is 28.62 meters.

ANSWER:

28.62 meters