Question

A ball was thrown from the top of a cliff. It reached a maximum height of 64 feet two seconds after it was thrown, and it hit the ground four seconds after it was thrown. How tall was the cliff?

Answer 1

EXPLANATION

Let's see the facts:

Maximum height = 64 feet Time = 2 seconds

Time to hit the ground = 4 seconds

Gravity = 9.8 m/s^2

The position equation is as follows:

`\text{Position}=(1)/(2)\cdot aceleration\cdot time^2+velocity\cdot time+initial\text{ position}`

Representing this situation:

Time to reach peak=

`velocity\text{ }+\text{ aceleration}\cdot time=0`

Replacing terms:

`initial\text{ velocity-}9,8\cdot2=0`

Isolating the initial velocity:

`\text{Initial velocity=9}.8(m)/(s^2)\cdot2s=19.6(m)/(s)`

Now, we now that at first segment, the ball travel 64 feet to the maximum height, thus we have a height of 64 feet. Now, we need to compute the distance to the ground applying the free fall kinematic equation.

`\text{Distance traveled from the top=}(1)/(2)\cdot g\cdot t^2`

As the time elapsed from the top was 4 seconds, the equation would be:

`\text{Distance trave}led\text{ from the top=}(1)/(2)\cdot9,8\cdot4^2`

Multiplying numbers and computing the powers:

`\text{Distance traveled from the top=}78.4\text{feet}`

Now, we need to subtract the distance traveled from the top to the maximum height reached in order to obtain the height:

`\text{Height of the cliff=Distance traveled from the top-Max}imum\text{ height reached}`

Replacing terms:

`\text{Height of the cliff=78.4 f}eet-64\text{ fe}et\text{ = 14.4 fe}et`

The cliff is 14.4 feet tall