1. EXAMPLE 4 Derive a trigonometric model Soccer Write an equation for the horizontal distance traveled by a soccer ball kicked from ground level (h 0 = 0) at speed v and angle .

2. EXAMPLE 4 Derive a trigonometric model SOLUTION Let h 0 = 0. 16 v 2 cos 2  x 2 + (tan  )x + 0 – = 0 Factor. – x( 16 v 2 cos 2  x – tan  ) = 0 Zero product property 16 v 2 cos 2  x – tan  = 0 Add tan  to each side. 16 v 2 cos 2  x = tan  Multiply each side by 1 16 v 2 cos 2 . x 1 16 v 2 cos 2  tan  =

3. EXAMPLE 4 Derive a trigonometric model Use cos  tan  = sin . x 1 16 v 2 cos  sin  = Rewrite as 2. 1 16 1 32 x 1 v 2 (2cos  sin  = Use a double-angle formula. x 1 32 v 2 sin 2  =

4. GUIDED PRACTICE for Examples 4 9. WHAT IF? Suppose you kick a soccer ball from ground level with an initial speed of 70 feet per second. Can you make the ball travel 200 feet? ANSWER No REASONING: Use the equation x = v 2 sin 2  to explain why the projection angle that maximizes the distance a soccer ball travels is  = 45°. 10. ANSWER sin 2(45°) = sin 90° = 1, this is the only value that will not decrease the velocity of the ball since it is the only non-fractional value. 1 32