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:2
Step-by-step explanation:
Answer: C. 120
Step-by-step explanation:
14/21 will get you 66.6. Taking that number and multiplying it by the 180 total students will end you with 120 which is your answer.
Answer:
P(t)=25000+1.12t
Step-by-step explanation:
Now can you help me?
A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following function:
f(t) = −16t2 + 16t + 32
Which of the following is a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground?
The Answer is (D. No solution)