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
Answer:
a > m/bp
Step-by-step explanation:
flip the equation
abp > m
divide both sides with bp
abp/bp > m/bp
a > m/bp
60 minutes in an hour
60÷12=5
1,000 meter in 1 km
5×1=5
5×1000=5000
therefore, 5000 meters
Answer:
The maximum height of the prism is
Step-by-step explanation:
Let
x------> the height of the prism
we know that
the area of the rectangular base of the prism is equal to
so
-------> inequality A
------> equation B
-----> equation C
Substitute equation B in equation C
------> equation D
Substitute equation B and equation D in the inequality A
-------> using a graphing tool to solve the inequality
The solution for x is the interval---------->
see the attached figure
but remember that
The width of the base must be meters less than the height of the prism
so
the solution for x is the interval ------>
The maximum height of the prism is