A projectile propelled straight upward from the ground is modeled by the function d(t)=-16t^2+304t. a) At what time does the projectile reach maximum height? b) What is the maximum height? c) At what times is the projectile more than 500 feet above the ground. Write your answer using interval notation. d)At what time does the projectile reach the ground?
(a) as with all parabolas, max height at t = -b/2a = 304/32
(b) max height is d(304/32)
(c) find the roots of d(t)=500. d(t)>500 between the roots.
(d) hits ground at d(t) = 0. t=304/16
Don't forget what you know about parabolas from Algebra I.
a) To find the time at which the projectile reaches maximum height, we need to determine the vertex of the parabolic function. The equation of the projectile's height is modeled by the function d(t) = -16t^2 + 304t.
The vertex of a parabola in the form of ax^2 + bx + c is given by (-b/2a , f(-b/2a)), where f(x) represents the function. In this case, a = -16 and b = 304.
Using the formula, the time at which the projectile reaches maximum height is t = -b/2a.
Plugging in the values, we get t = -304 / (2 * -16) = -304 / -32 = 9.5 seconds.
Therefore, the projectile reaches its maximum height at t = 9.5 seconds.
b) To find the maximum height, substitute the time found in part a (t = 9.5) into the function d(t) = -16t^2 + 304t.
d(9.5) = -16(9.5)^2 + 304(9.5)
= -16(90.25) + 2888
= -1444 + 2888
= 1444 feet
Hence, the maximum height of the projectile is 1444 feet.
c) To find the times at which the projectile is more than 500 feet above the ground, we set the function d(t) above 500 and solve for t.
-16t^2 + 304t > 500
Simplifying further, we get:
16t^2 - 304t + 500 < 0
Using the quadratic formula, which states that the roots are given by (-b ± √(b^2 - 4ac)) / (2a), we can determine the intervals.
The discriminant, b^2 - 4ac, is equal to (-304)^2 - 4(16)(500) = 92416 - 32000 = 60416, which is positive.
Since the coefficient of t^2 is positive, the parabola opens downward. Thus, the function is greater than zero between the roots.
Using the quadratic formula:
t = (-(-304) ± √(60416)) / (2*16)
t = (304 ± √60416) / 32
Calculating the roots:
t ≈ (304 + 246.97) / 32 ≈ 18.61 seconds
t ≈ (304 - 246.97) / 32 ≈ 1.8 seconds
Therefore, the projectile is more than 500 feet above the ground between 1.8 and 18.61 seconds.
Writing this in interval notation, the answer is [1.8, 18.61].
d) To find when the projectile reaches the ground, we set the function d(t) equal to zero and solve for t.
-16t^2 + 304t = 0
Factoring out common terms:
-16t(t - 19) = 0
Using the zero product property, we have two solutions:
t = 0 or t = 19
Therefore, the projectile reaches the ground at t = 0 seconds and t = 19 seconds.
In summary:
a) The projectile reaches maximum height at t = 9.5 seconds.
b) The maximum height of the projectile is 1444 feet.
c) The projectile is more than 500 feet above the ground between 1.8 and 18.61 seconds: [ 1.8, 18.61 ].
d) The projectile reaches the ground at t = 0 seconds and t = 19 seconds.