Exercise 46 Page &nbsp; 267

t= 0.81

0.81=0.5sqrt(h)

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Solve for h

LHS * 2=RHS* 2

1.62=sqrt(h)

LHS^2=RHS^2

2.6244=h

Rearrange equation

h=2.6244

Round to 2 decimal place(s)

h≈ 2.62

The jump height of a kangaroo is about 2.62 feet.

Snowboarder

t=0.5sqrt(h)

t= 1.21

1.21=0.5sqrt(h)

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Solve for h

LHS * 2=RHS* 2

2.42=sqrt(h)

LHS^2=RHS^2

5.8564=h

Rearrange equation

h=5.8564

Round to 2 decimal place(s)

h≈ 5.86

The jump height of a snowboarder is about 5.86 feet.

To double the hang times, we multiply the times given in Part A by 2.

hang time kangaroo:& 0.81(2)=1.62 s hang time snowboarder:& 1.21(2)=2.42 sNow we can calculate the new heights of the kangaroo and snowboarder.

Kangaroo

t=0.5sqrt(h)

t= 1.62

1.62=0.5sqrt(h)

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Solve for h

LHS * 2=RHS* 2

3.24=sqrt(h)

LHS^2=RHS^2

10.4976=h

Rearrange equation

h=10.4976

Round to 2 decimal place(s)

h≈ 10.50

The height of the kangaroo is about 10.5 feet when the hang time from Part A is doubled.

Snowboarder

t=0.5sqrt(h)

t= 2.42

2.42=0.5sqrt(h)

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Solve for h

LHS * 2=RHS* 2

4.84=sqrt(h)

LHS^2=RHS^2

23.4256=h

Rearrange equation

h=23.4256

Round to 2 decimal place(s)

h≈ 23.43

The height of the snowboarder is about 23.43 feet when the hang time from Part A is doubled.

To investigate how the jump increases when hang time doubles, we can divide the height when the time is T and compare it with 2T, which is double that of T.

Time is T

t=0.5sqrt(h)

t= T

T=0.5sqrt(h)

▼

Solve for h

LHS * 2=RHS* 2

2T=sqrt(h)

LHS^2=RHS^2

(2T)^2=h

PowProdII

(a b)^m=a^m b^m

4T^2=h

Rearrange equation

h=4T^2

The height when the time is T is 4T^2

Time is 2T

t=0.5sqrt(h)

t= 2T

2T=0.5sqrt(h)

▼

Solve for h

LHS * 2=RHS* 2

4T=sqrt(h)

LHS^2=RHS^2

(4T)^2=h

PowProdII

(a b)^m=a^m b^m

16T^2=h