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 for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter13: Conic Sections
Section13.1: Circles
Problem 48PS
Related questions
Question
For part (a) how did you get the answer 1,5 cause I get stuck on the part
Transcribed Image Text:11:18
when s = 80ft, putting this in the equation
gives,
80 = - 16t² + 96t
⇒16t² - 96t + 80 = 0
and we get,
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 =
16² +96
⇒ 16t²
-
96t
⇒ t = 6s
Thus the particle reaches the ground at time 6
√x
<
:
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