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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