Kayden is throwing a shot put at a track meet. He is 4.1

Question

feet tall. The shot put leaves Kayden's hand at a height of 6 feet. It moves with an upward velocity of 48 feet per second. Using the function h=−16t2+48t+6
, which of the following correctly identifies the start and the vertex, plus also explains its meaning? Round to the nearest hundredth if necessary.(1 point)
Responses

(4.1,0);
(42,1.5)
; The shot put is thrown from a distance of 0
feet at a height of 4.1
feet and will land after 42
seconds.
left parenthesis 1.19 comma 22.56 right parenthesis ; The shot put is thrown from a distance of 0 feet at a height of 4 point 1 feet and will land after 42 seconds.

(0,4.1);
(1.5,42)
; The shot put is thrown at a height of 0
feet and a distance of 4.1
feet and will reach a maximum height of 1.5
feet 42
seconds after being thrown.
left parenthesis 1.19 comma 22.56 right parenthesis ; The shot put is thrown at a height of 0 feet and a distance of 4 point 1 feet and will reach a maximum height of 1 point 5 feet 42 seconds after being thrown.

(6,0);
(42,1.5)
; The shot put is thrown at a height of 6
feet. It will reach a maximum height of 42
feet 1.5
seconds after being thrown.
left parenthesis 22.56 comma 1.19 right parenthesis ; The shot put is thrown at a height of 6 feet. It will reach a maximum height of 42 feet 1 point 5 seconds after being thrown.

(0,6);
(1.5,42)
; The shot put leaves Kayden's hand at a distance of 0
feet and a height of 6
feet and will reach a maximum height of 42
feet 1.5
seconds after being thrown.
left parenthesis 1.19 comma 22.56 right parenthesis ; The shot put leaves Kayden's hand at a distance of 0 feet and a height of 6 feet and will reach a maximum height of 42 feet 1 point 5 seconds after being thrown.
Skip to navigation
page 17 of 17

Answer 1

To analyze the function \( h = -16t^2 + 48t + 6 \) that describes the height of the shot put over time, we need to determine the initial height (start point) and the vertex of the parabola.

1. Start Point: The height at \( t = 0 \) (the time the shot put is thrown) can be found by substituting \( t = 0 \) into the function: \[ h(0) = -16(0)^2 + 48(0) + 6 = 6 \text{ feet} \] So, the shot put starts at \( (0, 6) \).

2. Vertex: The vertex of the parabola, which represents the maximum height of the shot put, can be found using the vertex formula for a quadratic function \( h(t) = at^2 + bt + c \): \[ t = -\frac{b}{2a} \] Here, \( a = -16 \) and \( b = 48 \): \[ t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5 \text{ seconds} \]

   To find the maximum height (the vertex), substitute \( t = 1.5 \) back into the height function: \[ h(1.5) = -16(1.5)^2 + 48(1.5) + 6 \] \[ = -16(2.25) + 72 + 6 \] \[ = -36 + 72 + 6 = 42 \text{ feet} \]

So, the vertex is at \( (1.5, 42) \).

Putting this all together, the start point is \( (0, 6) \) and the vertex is \( (1.5, 42) \). This means the shot put leaves Kayden’s hand at a height of 6 feet and reaches a maximum height of 42 feet after 1.5 seconds.

Based on these calculations, the correct option is:

(0,6); (1.5,42); The shot put leaves Kayden's hand at a distance of 0 feet and a height of 6 feet and will reach a maximum height of 42 feet 1.5 seconds after being thrown.