   Chapter 3.9, Problem 61E

Chapter
Section
Textbook Problem

# An object is projected upward with initial velocity v 0 meters per second from a point s 0 meters above the ground. Show that [ v ( t ) ] 2 = v 0 2 − 19.6 [ s ( t ) − s 0 ]

To determine

To show:

[vt]2=v02-19.6st-s0

Explanation

1) Concept:

i. If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is Fx+c, where c is an arbitrary constant.

ii.

αt=ddxvt

iii.

vt=ddxst

2) Given:

An object is projected upward with the initial velocity v0 meters per second from a point s0 meters above the ground.

3) Calculations:

Here, an object is projected upward with the initial velocity v0 meters per second from a point s0 meters above the ground.

Acceleration is equal to the force of gravity on earth, which is 9.8 m/s

That is, α=-9.8m/s because the object is falling.

Take general antiderivative of αt by using the rules of antiderivative,

vt=-9.8t+v0, using v0(initial velocity) as a constant term.

Now, take general antiderivative of vt using the rules of antiderivative,

st=-9.8t22+v0t+s0, using s0(initial position) as a constant term.

st=-4.9t2+v0t+s0

So, now taking squaring of vt we get,

vt2=-9.8t+v02

vt2=-9.82t2-2*9.8*v0+v02

vt2=9.8*9

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 