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: Understand the height function

The height of the rock above the water is given by the function \( h = -4.9(t - 2)^2 + 29 \). To find the initial height (when the rock is tossed, \( t = 0 \)), we substitute \( t = 0 \) into the function.

Step2: Substitute \( t = 0 \)

Substitute \( t = 0 \) into \( h = -4.9(t - 2)^2 + 29 \):

$$
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$$

Step1: Analyze the quadratic function

The height function \( h=-4.9(t - 2)^2+29 \) is a quadratic function in the form \( y = a(x - h)^2 + k \), where \( a=-4.9\), \(h = 2\) and \(k = 29\). For a quadratic function \( y=a(x - h)^2+k\), if \(a<0\), the parabola opens downwards and the vertex \((h,k)\) gives the maximum point.

Step2: Determine maximum height and time

Since \(a=-4.9<0\), the vertex of the parabola is the maximum point. The vertex of the function \(h=-4.9(t - 2)^2+29\) is at \((t = 2,h = 29)\). So the maximum height of the rock is \(29\) metres and it is reached at \(t = 2\) seconds.

Step1: Analyze the time when the rock hits the water

The rock hits the water when \(h = 0\). We set \(h=-4.9(t - 2)^2+29=0\) and solve for \(t\):

$$
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$$

We take the positive root (since time cannot be negative in this context): \(t=2+\sqrt{5.92}\approx2 + 2.43\approx4.43\) seconds.

Step2: Compare with \(t = 4.5\) seconds

Since \(4.5>4.43\), the rock has already hit the water (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)