   Chapter 5.1, Problem 66E ### 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.A ball is thrown upward with an initial velocity of 60 feet per second from an initial height of 16 feet. Express the height s (in feet) of the ball as a function of the time t (in seconds). How long will the ball be in the air?

To determine

To calculate: The height s (in feet) of the ball as a function of time t (seconds) which is thrown with an initial velocity of 60feetpersecond from an initial height of 16 feet and also calculate the time for which the ball is in the air.

Explanation

Given Information:

The initial velocity of the ball is 60feetpersecond and the initial height of the ball through which the ball is thrown is 16 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 ball thrown upward in the air for time t 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 ball at time t.

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)=60

Substitute 60 for s'(0) in above function for initial velocity condition.

60=0+C1C1=60

Therefore, the value of constant C1 is 60.

Substitute 60 for C1 in velocity function s'(t)=32t+C1.

s'(t)=32t+62

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

s'(t)dt=(32t+60)dts(t)=32(t1+11+1)+60t+C2=32(t22)+60t+C2=16t2+60t+C2

Substitute 0 for t in above height function for initial condition.

s(0)=16(0)2+60(0)+C2s(0)=C2

Substitute 16 for s(0) in above height function for initial height condition.

16=C2C2=16

Therefore, the value of constant C2 is 16.

Substitute 16 for C2 in height function s(t)=16t2+60t+C2

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