An object is launched from certain height into the air. The height is modeled by the the function h(t) = -4.8t^2+23t+20. The time (t) is measured in seconds.

A) What is the maximum height?
B) After how many seconds will it hit the ground?
C) From what height (initial height) was the object launched?
D) determine the domain for the above scenario?

recall that the vertex (maximum height) of at^2+bt+c is at t = -b/2a

it hits the ground when h=0, so just solve the quadratic as usual.

h(0) = 20

while the function has domain (-∞,+∞), we have not yet achieved time travel, so in the real world, the domain is [0,∞)

Going back in time, when h=0, that would be how long ago a missile would have to be fired from ground level to achieve the same flight path.

A) I have the maximum height 47.55 or 47.6

B) I have 5.5 seconds that it will hit the ground

Are these correct? Thanks

10.1 therom

Lilly, you are correct.

"therom," you are out of order.

To find the maximum height, we need to find the vertex of the parabolic function h(t) = -4.8t^2 + 23t + 20. The vertex of a parabola is the highest or lowest point on the graph.

A) To find the maximum height, we can use the formula for the x-coordinate of the vertex, which is given by the formula t = -b / (2a), where a, b, and c are the coefficients of the quadratic function. In this case, a = -4.8 and b = 23.

t = -23 / (2 * -4.8)
t ≈ 2.3958 seconds

To find the value of the function at this time, we substitute t = 2.3958 into the equation:

h(2.3958) = -4.8(2.3958)^2 + 23(2.3958) + 20
h(2.3958) ≈ 42.9562

Therefore, the maximum height is approximately 42.9562 units.

B) To find when the object hits the ground, we need to find the time t when the height is equal to zero. So we set h(t) = 0 and solve for t.

-4.8t^2 + 23t + 20 = 0

This equation can be solved using factoring, completing the square, or the quadratic formula. In this case, we will use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values into the formula:

t = (-23 ± √(23^2 - 4 * -4.8 * 20)) / (2 * -4.8)
t ≈ 4.042 seconds or t ≈ 0.524 seconds

Since time cannot be negative, we ignore the negative value. So, after approximately 4.042 seconds, the object will hit the ground.

C) To determine the initial height from which the object was launched, we need to find h(0), which represents the height at time t = 0.

h(0) = -4.8(0)^2 + 23(0) + 20
h(0) = 20

Therefore, the object was launched from a height of 20 units.

D) The domain of the scenario is the set of all possible input values for the function, which in this case is time t. Since time cannot be negative (as it represents the time after launch), the domain for this scenario is t ≥ 0.