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
The answer is 2 to the power of 6
Answer:
x= 19/7
Step-by-step explanation:
I don’t have a calculator with me use sine 20divided by Sin(60) will give you x
Answer:
The answers would be 9 and 1.
Step-by-step explanation:
x= 5±√16 can be separated into two equations.
x=5+√16 and x=5-√16
1. 5+√16 can be solved by simplify √16, which is 4. So then it would become 5+4 which equals 9.
2. 5-√16 can be solved by simplifying √16 which is 4. So then it would become 5-4 which equals 1.
So the values of x are 9 and 1.