# A silver dollar is dropped from a building that is 1,362 feet in height. What is the velocity?

## Question:

A silver dollar is dropped from a building that is 1,362 feet in height. What is the velocity?

## Acceleration due to Gravity:

When air resistance is neglected and a body is dropped near the surface of the earth it falls along a vertical straight line with a constant acceleration that is called the acceleration due to gravity on the earth, is denoted by *g*. Suppose the body moves upward then the acceleration of the body is negative {eq}a = - g {/eq} and moves downward then the acceleration is positive where {eq}g = 9.8 \ \rm m / s^2 \text{ or } 32.2 \ \rm ft / s^2 {/eq}

The scientist gave the motion equations when the acceleration of the body is constant that are given by: {eq}v^2 = u^2 + 2 \ a \ s {/eq}

## Answer and Explanation:

Given:

- The height of the building is: {eq}s = 1,362 \ \rm ft {/eq}.

As we know when an object is dropped from the some height at which the initial velocity is zero, so we can write {eq}u = 0 {/eq}

We will compute the velocity when the silver dollar strikes the ground.

Apply the Newton's motion equation:

$$\begin{align*} \displaystyle v^2 &= u^2 + 2 \ a \ s &\text{(Where } s = 1,362 \ \rm ft , a = g = 32.2 \ \rm ft / s^2 \text{ and } u = 0 \text{)}\\ v^2 &= 0^2 + 2 \times 32.2 \times 1,362 \\ v^2 &= 0 + 87,712.8 \\ v^2 &= 87,712.8 \\ v &= \sqrt{87,712.8} &\text{(Square root both sides)}\\ v \ &\boxed{= 296.16 \ \rm ft / s} \end{align*} $$

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from General Studies Science: Help & Review

Chapter 18 / Lesson 5