For part (a) how did you get the answer 1,5 cause I get stuck on the part **Example Calculation of Projectile Motion**### Step 2The height of the projectile is given as:\[ s = -16t^2 + 96t \]When \( s = 80 \) ft, putting this in the equation gives:\[ 80 = -16t^2 + 96t \]\[ \Rightarrow 16t^2 - 96t + 80 = 0 \]Thus, the particle reaches the height of 80 ft at 1 second while it goes up and after 5 seconds while it moves downwards.### Step 3#### Part (b)When the particle reaches the ground, the height will be zero, and so solving the equation with \( s = 0 \) gives:\[ s = -16t^2 + 96t \]\[ \Rightarrow 0 = -16t^2 + 96t \]\[ \Rightarrow 16t^2 = 96t \]\[ \Rightarrow t = 6 \text{s} \]Thus, the particle reaches the ground at time 6 seconds.

Answered: Image /qna-images/answer/a875daa9-aed4-471a-86ca-d1f5f020eff1.jpg

Thus the particle reach the height of 80f at 1second while it goes up and after 5s while moves downwards. Step3 c) (b) When the particle reaches the ground, the height will be zero, and so solving the equation with s = 0 gives: S = 16t² + 96t 0 = 16t² + 96t ⇒ 16t² = 96t ⇒ t = 6s Thus the particle reaches the ground at time 6 vx Do

Thus the particle reach the height of 80f at 1second while it goes up and after 5s while moves downwards. Step3 c) (b) When the particle reaches the ground, the height will be zero, and so solving the equation with s = 0 gives: S = 16t² + 96t 0 = 16t² + 96t ⇒ 16t² = 96t ⇒ t = 6s Thus the particle reaches the ground at time 6 vx Do

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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