Answer:
10
Step-by-step explanation:
Ground level is where h = 0, so solve the equation ...
h(x) = 0
-5(x -4)^2 +180 = 0 . . . . substitute for h(x)
(x -4)^2 = 36 . . . . . . . . . . divide by -5, add 36
x -4 = 6 . . . . . . . . . . . . . . positive square root*
x = 10 . . . . . . add 4
The object will hit the ground 10 seconds after launch.
_____
* The negative square root also gives an answer that satisfies the equation, but is not in the practical domain. That answer would be x = -2. The equation is only useful for time at and after the launch time: x ≥ 0.
200 because:
250 or higher is 300
249 or lower is 200
Answer:
Step-by-step explanation:
I am sorry but what is this suppose to be asking
Answer:
62.5%
Step-by-step explanation:
plz mark the brainliest
Answer:
x₁ > x₂
Step-by-step explanation:
Both actions imply a parable trajectories, since both are projectile shot cases.
Let´s call x₁ maximum distance in the first case
The maximum height is just in the middle of the curve, therefore x₁ the maximum horizontal distance is equal to 60 feet.
In the second case, the parable curve is modeled by:
y = x₂*( 0.08 - 0.002x₂) or y = 0.08*x₂ - 0.002*x₂²
A second degree equation, solving for x₂ and dismissing the value x₂ = 0
we get:
y = 0 ⇒ x₂*( 0.08 - 0.002x₂) = 0 x₂ = 0
And 0,08 - 0.002*x₂ = 0
- 0.002*x₂ = - 0.08
x₂ = 0.08/0.002
x₂ = 40 f
Then x₁ > x₂
