   Chapter 5.1, Problem 68E ### 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
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# Vertical Motion In Exercises 65–68, use s′′(t) = −32 feet per second per second as the acceleration due to gravity. (Neglect air resistance.) See Example 7.With what initial velocity must an object be thrown upward from a height of 5 feet to reach a maximum height of 230 feet?

To determine

To calculate: The initial velocity requires to throw an object upward from a height of 5 feet to reach a maximum height of 230 feet.

Explanation

Given Information:

The initial height from the ball is 5 feet and the maximum height attain by the ball is 230 feet.

Formula used:

The constant rule of integration, kdx=kx+C.

The simple power rule of integration, xndx=xnn+1+C.

Calculation:

Consider the height of the object thrown upward is s.

The acceleration function due to gravity in downward direction is 32feet per second per second.

s''(t)=32

Integrate both sides, use the constant rule of integration kdx=kx+C.

s''(t)dt=32dts'(t)=32t+C1

Here, s'(t) is the velocity of the object at time t.

Consider the initial velocity of the object is v0.

Substitute 0 for t in the velocity function s'(t)=32t+C1 for initial velocity condition.

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

The initial velocity of the ball thrown upward is given as,

s'(0)=v0

Substitute v0 for s'(0) in above function for initial velocity condition.

v0=0+C1C1=v0

Substitute v0 for C1 in velocity function s'(t)=32t+C1

s'(t)=32t+v0

Integrate both side the above velocity function, use the simple power rule of integration xndx=xnn+1+C

s'(t)dt=(32t+v0)dts(t)=32(t1+11+1)+v0t+C2=32(t22)+v0t+C2=16t2+v0t+C2

The initial height of the ball is 5 feet

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