# The distance of stone above the ground level. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4.8, Problem 43E

(a)

To determine

## To find: The distance of stone above the ground level.

Expert Solution

The distance of stone above ground level at time t is 4504.9t2_ .

### Explanation of Solution

Given data:

The distance between upper observation deck and ground is 450 m.

Formula used:

Write the expression for acceleration function a(t) .

a(t)=v(t) (1)

Here,

v(t) is first derivative of velocity function v(t) .

Write the expression for velocity function v(t) .

v(t)=s(t) (2)

Here,

s(t) is first derivative of position function s(t) .

Antiderivative of t3 is 14t4 , t2 is 13t3 , t is 12t2 and 1 is t .

Calculation:

At first observation, velocity is zero and position is at 450 m at t=0 , as the stone is dropped 450 m height to ground level. Hence,

v(0)=0s(0)=450

The motion of stone is close to ground, so the motion is considered as gravitational constant (g) , which is 9.8ms2 .

Write the expression for acceleration function (a(t)) .

a(t)=9.8

Substitute 9.8 for a(t) in equation (1),

v(t)=9.8

Antiderivate the expression with respect to t,

v(t)=9.8t+C (3)

Here,

C is arbitrary constant.

Substitute 0 for t in equations (3),

v(0)=9.8(0)+C

v(0)=C (4)

Substitute 0 for v(0) in equation (4),

C=0

Substitute 0 for C in equation (3),

v(t)=9.8t+0

v(t)=9.8t (5)

Substitute 9.8t for v(t) in equation (2),

s(t)=9.8t+C

Antiderivate the expression with respect to t.

s(t)=9.8(12t2)+D

s(t)=4.9t2+D (6)

Here,

D is arbitrary constant.

Substitute 0 for t in equation (6),

s(0)=4.9(02)+D=0+D=D

Substitute 450 for s(0) ,

D=450

Substitute 450 for D in equation (6),

s(t)=4.9t2+450=4504.9t2

Thus, the distance of stone above ground level at time t is 4504.9t2_ .

(b)

To determine

### To find: The time that stone takes to reach ground.

Expert Solution

The time that stone takes to reach ground is 9.58 seconds.

### Explanation of Solution

Given data:

The distance between upper observation deck and ground is 450 m.

Calculation:

When the stone reaches the ground, the positional function reaches zero. Hence,

s(t)=0 (7)

Substitute 4504.9t2 for s(t) in equation (7),

4504.9t2=04.9t2=450t2=4504.9t2=91.84

Take square root on both sides.

t=91.84secondst9.58seconds

Thus, the time that stone takes to reach ground is 9.58 seconds.

(c)

To determine

### To find: The velocity that stone strikes the ground.

Expert Solution

The velocity that stone strikes the ground is 93.9ms_ .

### Explanation of Solution

Given data:

The distance between upper observation deck and ground is 450 m.

Calculation:

Substitute 9.58 seconds for t in equation (5),

v(t)=(9.8ms2)(9.58seconds)=93.88ms93.9ms

Thus, the velocity that stone strikes the ground is 93.9ms_ .

(d)

To determine

### To find: The time that stone takes to reach ground.

Expert Solution

The time that stone takes to reach ground, when it is thrown downward with speed of 5ms is 9.09 seconds.

### Explanation of Solution

Given data:

The speed is 5ms .

Formula used:

Write the expression to find the roots of at2+bt+c=0 .

t=b±b24ac2a (8)

Here,

a, b, and c are constants.

Calculation:

The stone is thrown downwards with a speed of 5ms . Hence v(0)=5 .

Substitute –5 for v(0) in equation (4),

C=5

Substitute –5 for C in equation (3),

v(t)=9.8t5=(9.8t+5)

Substitute (9.8t+5) for v(t) in equation (2),

s(t)=(9.8t+5)+C

Antiderivate the expression with respect to t,

s(t)=9.8(12t2)5t+D

s(t)=4.9t25t+D (9)

Here,

D is arbitrary constant.

Substitute 0 for t in equation (9),

s(0)=4.9(02)5(0)+D=0+D=D

Substitute 450 for s(0) ,

D=450

Substitute 450 for D in equation (9),

s(t)=4.9t25t+450=4504.9t25t

Substitute 4504.9t25t for s(t) in equation (7),

4504.9t25t=04.9t2+5t450=0

Compare the expressions ax2+bx+c=0 and 4.9t2+5t450=0 .

a=4.9b=5c=450

Substitute 4.9 for a, 5 for b, and –450 for c in equation (8),

t=5±524(4.9)(450)2(4.9)=5±25+88209.8=5±88459.8=5±94.059.8

Hence, the two possible values of t are,

t=5+94.059.8=89.059.8=9.0869.09seconds

t=594.059.8=99.059.8=10.01seconds

The time cannot be a negative, so the value of t is 9.09 seconds.

Thus, the time that stone takes to reach ground, when it is thrown downward with speed of 5ms is 9.09 seconds.

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