Question

lisc - shaped target is launched straight up into the sky. the targets height ( h ) (in meters) above the launcher after ( t ) seconds is given by the function ( h(t) = - 4.9t^{2}+14.5t ). an archer is high up on a nearby tower, 16 meters above the target launcher. she tracks the target using a sight as it travels into the sky. the horizontal distance between the archers eyes and the targets vertical path is 17 meters, as shown below. (the figure is not drawn to scale.)

Explanation:

Since the problem isn't fully stated (the actual question part is cut off), we can infer it might be about finding when the target's height plus 16 (the archer's height above the launcher) relates to the 17m distance, maybe finding \( t \) when the vertical height from the archer's eyes is \( h(t)+16 \), and using trigonometry (since there's a right triangle with hypotenuse 17m, horizontal 16m? Wait, no, the horizontal distance between the archer's eyes and the target's vertical path is 16m? Wait, the diagram: archer is 16m above the target launcher, and the horizontal distance between archer's eyes and target's vertical path is 17m? Wait, the text says "the horizontal distance between the archer’s eyes and the target’s vertical path is 17 meters" and "An archer is high up on a nearby tower, 16 meters above the target launcher". The target's height above the launcher is \( h(t) = -4.9t^2 + 14.5t \). So the vertical distance from the archer's eyes to the target is \( |h(t) - 16| \) (wait, no: archer is 16m above the launcher, target is \( h(t) \) above the launcher, so if target is above launcher, the vertical distance between archer and target is \( |16 - h(t)| \) if target is below archer, or \( h(t) - 16 \) if target is above. Then with horizontal distance 17m, maybe finding \( t \) when the angle \( \theta \) has some condition, or when the distance is a certain value, or when the target is at the same height as archer, or when the line of sight has a certain slope. But since the question is cut off, we can't proceed fully. But assuming the question is to find when the target is at the same height as the archer's eyes? Wait, archer is 16m above launcher, target's height above launcher is \( h(t) \), so set \( h(t) = 16 \):

Step1: Set up the equation

\( -4.9t^2 + 14.5t = 16 \)

Step2: Rearrange to standard quadratic form

\( 4.9t^2 - 14.5t + 16 = 0 \)

Step3: Use quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -14.5 \), \( c = 16 \)

First, calculate discriminant \( D = (-14.5)^2 - 4 \times 4.9 \times 16 \)
\( D = 210.25 - 313.6 = -103.35 \)
Wait, discriminant is negative, so no real solution? That can't be. Maybe the vertical distance is \( 16 + h(t) \)? Wait, no, archer is 16m above launcher, target is \( h(t) \) above launcher, so the vertical distance from archer to target is \( |h(t) - 16| \) if archer is 16m above launcher, target is \( h(t) \) above. Wait, maybe the archer is 16m below the target's vertical path? No, the text says "16 meters above the target launcher". So launcher is at ground, archer is 16m above ground, target is \( h(t) \) above ground. So vertical distance between archer and target is \( |h(t) - 16| \). Horizontal distance is 17m. Maybe the question is to find when the target is at the same height as the archer? So set \( h(t) = 16 \):

\( -4.9t^2 + 14.5t - 16 = 0 \)

Multiply both sides by -1: \( 4.9t^2 - 14.5t + 16 = 0 \)

Discriminant \( D = (14.5)^2 - 4 \times 4.9 \times 16 = 210.25 - 313.6 = -103.35 \), which is negative. So no real time when target is at archer's height. Maybe the question is about the angle \( \theta \), where \( \tan\theta = \frac{\text{vertical distance}}{\text{horizontal distance}} = \frac{|h(t) - 16|}{17} \), and find when \( \tan\theta \) is maximum or when the target is at the peak? The peak of \( h(t) \) is at \( t = -\frac{b}{2a} = -\frac{14.5}{2 \times (-4.9)} = \frac{14.5}{9.8} \approx 1.48 \) seconds, peak height \( h(1.48) = -4.9(1.48)^2 + 14.5(1.48) \approx -4.9(2.19) + 21.46 \approx -10.73 + 21.46 \approx 10.73 \) meters, which is less than 16m, so archer is always above the target? Wait, archer is 16m above launcher, target's max height is ~10.73m, so target never reaches archer's height. Maybe the question is to find when the distance between archer and target is a certain value. The distance \( d(t) = \sqrt{17^2 + (16 - h(t))^2} \) (since horizontal distance 17m, vertical distance \( 16 - h(t) \) because \( h(t) \leq 10.73 < 16 \), so vertical distance is \( 16 - h(t) \)). Maybe find \( t \) when \( d(t) \) is minimum or when \( 16 - h(t) = 17 \tan\theta \), but without the full question, we can't proceed. Since the original problem's question is cut off, we can't give a complete solution. But assuming the question was to find when the target is at the same height as the archer, the answer would be no real solution, but since the problem is incomplete, we note that.

Answer:

(The problem is incomplete; the actual question is missing. If assuming to solve \( -4.9t^2 + 14.5t = 16 \), discriminant is negative, so no real \( t \).)