Question

A ball is thrown vertically down from the edge of a cliff with a speed of 8 m/s, how high is the cliff, if it took 6 s for the ball to reach the ground?

Answer 1

You would use h=1/2*g*t^2
Therefore,
g= 9.81
h= 1/2*9.81*6^2
h= 176.58m

Answer 2

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

Given the data in the question;

• Initial velocity of the ball; $u = 8m/s$

• Time taken by the ball to reach the ground; $t = 6s$

• Distance or Height of the cliff from which the ball was thrown from; $s =\ ?$

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

$s = ut + \frac{1}{2}at^2$

Where s is the distance or height, $u$ 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; $g = 9.8m/s^2$

Hence, the equation becomes

$s = ut + \frac{1}{2}gt^2$

We substitute the given values into the equation

$s = ( 8m/s\ *\ 6s) + (\frac{1}{2}\ *\ 9.8m/s^2\ *\ (6s)^2)\\\\ s = ( 48m ) + (4.9m/s^2\ *\ 36s^2)\\\\ s = ( 48m ) + (4176.4m)\\\\ s = 224.4m$

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

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