Question

a disc - shaped target is launched straight up into the sky. the targets height ( h ) (in meters) above the launcher is given by ( 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. the distance between the archers eyes and the targets vertical path is 17 meters, as shown below. (the figure shows a right - angled triangle - like diagram with one side labeled 17 m and another related to the 16 m height of the archer above the launcher.)

Explanation:

Step1: Define the target's height

The target's height above the launcher is given by \( h(t) = -4.9t^2 + 14.5t \). The archer's eye height is 16 m above the launcher, so the vertical distance between the target and the archer's eyes is \( |h(t) - 16| \).

Step2: Set up the trigonometric relation

We know the horizontal distance (adjacent side) is 17 m, and the vertical distance is \( |h(t) - 16| \) (opposite side for angle \( \theta \)). Using the tangent function: \( \tan\theta=\frac{|h(t)-16|}{17} \).

Step3: Substitute \( h(t) \) into the formula

Substitute \( h(t)= -4.9t^2 + 14.5t \) into the tangent formula: \( \tan\theta=\frac{|-4.9t^2 + 14.5t - 16|}{17} \). If we want to find when the target is at the same height as the archer's eyes (i.e., \( h(t)=16 \)), we solve \( -4.9t^2 + 14.5t = 16 \), or \( 4.9t^2 - 14.5t + 16 = 0 \). Using the quadratic formula \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b=-14.5 \), \( c = 16 \). Calculate the discriminant: \( \Delta=(-14.5)^2 - 4\times4.9\times16 = 210.25 - 313.6=-103.35 \). Since the discriminant is negative, the target never reaches the archer's eye height (16 m) above the launcher. So the vertical distance is \( 16 - h(t) \) (since \( h(t) \) is always less than 16? Wait, no, let's check the vertex of \( h(t) \). The vertex of a parabola \( at^2+bt+c \) is at \( t = -\frac{b}{2a} \). For \( h(t)=-4.9t^2 + 14.5t \), \( t=\frac{14.5}{2\times4.9}\approx1.48 \) s. Then \( h(1.48)=-4.9\times(1.48)^2 + 14.5\times1.48\approx -4.9\times2.19 + 21.46\approx -10.73 + 21.46 = 10.73 \) m, which is less than 16 m. So the vertical distance is \( 16 - h(t) \), so \( \tan\theta=\frac{16 - (-4.9t^2 + 14.5t)}{17}=\frac{4.9t^2 - 14.5t + 16}{17} \). But maybe the question is to find \( \theta \) when? Wait, the problem is not fully stated, but assuming we need to find \( \theta \) in terms of \( t \), the formula is \( \tan\theta=\frac{16 - (-4.9t^2 + 14.5t)}{17}=\frac{4.9t^2 - 14.5t + 16}{17} \), or if we consider the target's height relative to the archer, but since the discriminant was negative, the target is always below the archer's eyes (in height above launcher), so the vertical distance is \( 16 - h(t) \), so \( \tan\theta=\frac{16 - h(t)}{17} \), substituting \( h(t) \) gives \( \tan\theta=\frac{16 - (-4.9t^2 + 14.5t)}{17}=\frac{4.9t^2 - 14.5t + 16}{17} \). If we want a numerical value, we need a specific \( t \), but since the problem is cut off, maybe this is the relation.

Answer:

The tangent of the angle \( \theta \) is given by \( \boldsymbol{\tan\theta=\frac{4.9t^2 - 14.5t + 16}{17}} \) (or if considering absolute value, but since \( h(t) < 16 \) for all real \( t \), this is valid).