856 CHAPTER 8 Applications of Trigonometry EXAMPLE 7 Examining Parametric Equations of Flight Jack launches a small rocket from a table that is 3.36 ft above the ground. Its initial velocity is 64 ft per sec, and it is launched at an angle of 30° with respect to the ground. Find the rectangular equation that models its path. What type of path does the rocket follow? SOLUTION The path of the rocket is defined by the parametric equations x = 164 cos 30°2t and y = 164 sin 30°2t - 16t2 + 3.36 or, equivalently, x = 3223 t and y = -16t2 + 32t + 3.36. Using x = 3223 t, we solve for t. t = x 3223 Divide by 3223. Substituting for t in the other parametric equation yields the following. y = -16t2 + 32t + 3.36 y = -16 ¢ x 3223≤ 2 + 32 ¢ x 3223≤ + 3.36 Let t = x 3223 . y = - 16x2 1024132 + x2 3 + 3.36 y = - 1 192 x2 + 23 3 x + 3.36 Simplify. This equation defines a parabola. The rocket follows a parabolic path. S Now Try Exercise 51(a). Use rules for exponents and multiply. EXAMPLE 8 Analyzing the Path of a Projectile Determine the total flight time and the horizontal distance traveled by the rocket in Example 7. ALGEBRAIC SOLUTION The equation y = -16t2 + 32t + 3.36 tells the vertical position of the rocket at time t. We need to determine the positive value of t for which y = 0 because this value corresponds to the rocket at ground level. This yields 0 = -16t2 + 32t + 3.36. Using the quadratic formula, the solutions are t = -0.1 or t = 2.1. Because t represents time, t = -0.1 is an unacceptable answer. Therefore, the flight time is 2.1 sec. The rocket was in the air for 2.1 sec, so we can use t = 2.1 and the parametric equation that models the horizontal position, x = 3223 t, to obtain x = 3223 12.12 ≈116.4 ft. −10 −30 50 150 Figure 82 S Now Try Exercise 51(b). GRAPHING CALCULATOR SOLUTION Figure 82 shows that when t = 2.1, the horizontal distance x covered is approximately 116.4 ft, which agrees with the algebraic solution.
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