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

(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.

(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.

(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.

(0,4.1);
(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.

Answer 1

To analyze the situation, let's break it down using the function \( h = -16t^2 + 48t + 6 \), which models the height \( h \) of the shot put over time \( t \).

1. Start Point: At time \( t = 0 \) seconds, we can find the height of the shot put. \[ h(0) = -16(0)^2 + 48(0) + 6 = 6 \text{ feet} \] So the shot put leaves Kayden's hand at a height of 6 feet.

2. Vertex: To find the time at which the shot put reaches its maximum height, we can use the formula for the vertex of a quadratic function \( t = -\frac{b}{2a} \) where \( a = -16 \) and \( b = 48 \). \[ t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5 \text{ seconds} \] Now, substituting \( t = 1.5 \) back into the equation to find the maximum height: \[ h(1.5) = -16(1.5)^2 + 48(1.5) + 6 \] Calculating gives: \[ h(1.5) = -16(2.25) + 72 + 6 = -36 + 72 + 6 = 42 \text{ feet} \] So the maximum height is 42 feet, reached at 1.5 seconds.

Based on this analysis, the correct description would identify the start point (0, 6) and the vertex (1.5, 42), explaining that 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.

The correct response 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.