The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.
Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s
The projectile reaches a height of 192 ft at 3 s on the way up, and at 4 s on the way down.
Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s
When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.
Answer: 7 s
I got C but not sure if that is right
Answer:
Step-by-step explanation:
624 students
Since they’re only 3 parts more girls then boys and the number equals 144, you just have to divide 144 by 3. You’ll get 48
Multiply 5 and 8 by 48 to get 240 and 384 and add to get 624 which is the answer
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The scale for the second values is 56/8 = 7
C = 3 x 7 = 21
C= 21