Question

1. a rock is tossed into the air from a bridge over a river. its height ( h ) above the water, in metres, after ( t ) seconds is ( h = -4.9(t - 2)^2 + 29 ).

(a) from what height above the water is the rock tossed?
(b) find the maximum height of the rock and the time when this height is reached.
(c) is the ball still in the air after 4.5 s?

Explanation:

Part (a)

Step1: Analyze the vertex form of the quadratic function

The height function is given in vertex form \( h = -4.9(t - 2)^2 + 29 \). In the vertex form of a quadratic function \( y = a(x - h)^2 + k \), the vertex is at \( (h, k) \). Here, when \( t = 0 \) (initial time, when the rock is tossed), we substitute \( t = 0 \) into the equation.
\( h = -4.9(0 - 2)^2 + 29 \)

Step2: Calculate the value

First, calculate \( (0 - 2)^2 = (-2)^2 = 4 \). Then, \( -4.9\times4=-19.6 \). Finally, \( h=-19.6 + 29 = 9.4 \).

Step1: Identify the vertex from the vertex form

The height function is \( h = -4.9(t - 2)^2 + 29 \), which is in the vertex form of a quadratic function \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. For a quadratic function with \( a<0 \) (here \( a=-4.9<0 \)), the vertex represents the maximum point.

Step2: Determine the maximum height and time

From the vertex form, the vertex is at \( (t, h)=(2, 29) \). So the maximum height \( k = 29 \) metres is reached at time \( t = 2 \) seconds.

Step1: Substitute \( t = 4.5 \) into the height function

We have the height function \( h=-4.9(t - 2)^2 + 29 \). Substitute \( t = 4.5 \) into the function:
First, calculate \( t - 2=4.5 - 2 = 2.5 \). Then, \( (t - 2)^2=(2.5)^2 = 6.25 \). Next, \( -4.9\times6.25=-30.625 \). Finally, \( h=-30.625+29=-1.625 \).

Step2: Analyze the height value

A negative height means the rock is below the water level (since height above water is positive). So the rock is not in the air after \( 4.5 \) seconds.

Answer:

The rock is tossed from a height of \( 9.4 \) metres above the water.

Part (b)