Question

A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s, how high is the cliff, if it took 12 s for the ball to reach the ground?
I need the Formula

Answer 1

Answer:

The height of the cliff from which the ball was dropped from is 224.4m.

\overline{v}={\frac{\Delta x}{\Delta t}}

Given the data in the question;

Initial velocity of the ball;

Time taken by the ball to reach the ground;

Distance or Height of the cliff from which the ball was thrown from;

To get the height of the Cliff, we use the Second Equation of Motion:

Where s is the distance or height,  is the initial velocity, t is the time and a is the acceleration. Since the ball was thrown down from a certain height (cliff), its is now under the influence of gravity. acceleration due to gravity;

Hence, the equation becomes

We substitute the given values into the equation

Therefore, the height of the cliff from which the ball was dropped from is 224.4m

Explanation:

Answer 2

Hi there!

We can use the equation:

d = v₀t + 1/2at²

Where:

v₀ = initial velocity downward

a = acceleration due to gravity

t = time

Plug in given values:

d = 4(12) + 1/2(9.8)(12²)

d = 48 + 705.6 = 753.6 m