   Chapter 5.1, Problem 7CP ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
1 views

# Checkpoint 7 Worked-out solution available at LarsonAppliedCalculus.comDerive the position function when a ball is thrown upward with an initial velocity of 32 feet per second from an initial height of 48 feet. When does the ball hit the ground? With what velocity does the ball hit the ground?

To determine

To calculate: The position function of a ball when it is thrown upward with an initial velocity 32feet per second from an initial height of 48feet and also estimate the time when the ball hit the ground and the hitting velocity of the ball at ground.

Explanation

Given Information:

The initial velocity of the ball which is thrown upward is 32feet per second and the initial height from the ground is 48feet. The acceleration due to gravity in downward direction is 32feet per second per second.

Formula used:

The basic rule of integration.

[f(x)+g(x)]dx=f(x)dx+g(x)dx

Calculation:

Consider the initial height (s(0)) 48feet and initial velocity of the ball (s'(0)) 32feet per second.

The acceleration due to gravity of the ball in upward direction is,

s''=32

Integrate both side with respect to time (t).

s''dt=32dts'(t)=32t+C1 …… (1)

Substitute 0 for t in above equation for initial condition.

s'(0)=32(0)+C1s'(0)=C1

Substitute 32 for s'(0) in above equation for initial velocity condition.

32=C1C1=32

Therefore, the value of constant C1 is 32.

Substitute 32 for C1 in equation (1).

s'(t)=32t+32 …… (2)

Integrate both side with respect to time (t).

s'dt=(32t+32)dts(t)=32(t22)+32t+C2=16t2+32t+C2 …… (3)

Substitute 0 for t in above equation for initial condition.

s(0)=16(02)+32(0)+C1s'(0)=C2

Substitute 48 for s(0) in above equation for initial height condition.

48=C2C2=8

Therefore, the value of constant C2 is 48.

Substitute 48 for C2 in equation (3).

s(t)=16t2+32t+48

For the time to hit the ground the position function s(t) should be zero.

Substitute 0 for s(t) in above equation.

0=16t2+32t+4816t2+32t+48=016(t22t3)=0

The term 16 cannot be zero so, the term t22t3 must be zero

### 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 