   Chapter 5.1, Problem 67E ### 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 ground level) to reach the top of the Washington Monument (555 feet)?

To determine

To calculate: The initial velocity requires to throw an object upward (from ground level) to the top of the Washington Monument (555 feet).

Explanation

Given Information:

The height of the Washington Monument from ground level to the top is 555 feet.

Formula used:

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

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

The zero property of multiplication, if ab=0 the either a=0 or b=0.

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

Substitute 0 for t in above height function for initial condition

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