EXAMPLE 5 Calculate horizontal distance traveled

Robotics

The “frogbot” is a robot designed for exploring rough terrain on other planets. It can jump at a 45° angle and with an initial speed of 16 feet per second. On Earth, the horizontal distance d (in feet) traveled by a projectile launched at an angle θ and with an initial speed v (in feet per second) is given by:

d = v2

32sin 2θ

How far can the frogbot jump on Earth?

EXAMPLE 5 Calculate horizontal distance traveled

SOLUTION

d = v2

32sin 2θ

d = 162

32sin (2 45°)

= 8

Write model for horizontal distance.

Substitute 16 for v and 45° for θ.

Simplify.

The frogbot can jump a horizontal distance of 8 feet on Earth.

EXAMPLE 6 Model with a trigonometric function

Rock climbing

A rock climber is using a rock climbing treadmill that is 10.5 feet long. The climber begins by lying horizontally on the treadmill, which is then rotated about its midpoint by 110° so that the rock climber is climbing towards the top. If the midpoint of the treadmill is 6 feet above the ground, how high above the ground is the top of the treadmill?

EXAMPLE 6 Model with a trigonometric function

SOLUTION

sin θ =yr

sin 110° =y

5.25

4.9 y

Use definition of sine.

Solve for y.

The top of the treadmill is about 6 + 4.9 = 10.9 feet above the ground.

Substitute 110° for θ and = 5.25 for r.210.5

GUIDED PRACTICE for Examples 5 and 6

10. Estimate the horizontal distance traveled by a track and field long jumper who jumps at an angle of 20° and with an initial speed of 27 feet per second.

TRACK AND FIELD

SOLUTION

d = v2

32sin 2θ

d = 272

32sin (2 20°)

= 14.64

Write model for horizontal distance.

Substitute 27 for v and 20° for θ.

Simplify.

The long jumper can jump 14.64 feet.

GUIDED PRACTICE for Examples 5 and 6

11. WHAT IF? In Example 6, how high is the top of the rock climbing treadmill if it is rotated 100° about its midpoint?

SOLUTION

sin θ =yr

sin 100° =y

5.25

5.2 y

Use definition of sine.

Solve for y.

The top of the treadmill is about 6 + 5.2 = 11.2 feet above the ground.

Substitute 100° for θ and = 5.25 for r.210.5