Solve the problem.
-A flare fired from the bottom of a gorge is visible only when the flare is above the rim. If it is fired with an initial velocity of $176 \mathrm { ft } / \mathrm { sec }$, and the gorge is $448 \mathrm { ft }$ deep, during what interval can the flare be seen?
$$\left( h = - 16 t ^ { 2 } + v _ { O } t + h _ { O } \right)$$
A) $8

Question

Solve the problem.
-A flare fired from the bottom of a gorge is visible only when the flare is above the rim. If it is fired with an initial velocity of $176 \mathrm { ft } / \mathrm { sec }$, and the gorge is $448 \mathrm { ft }$ deep, during what interval can the flare be seen?
$$\left( h = - 16 t ^ { 2 } + v _ { O } t + h _ { O } \right)$$
A) $8 < t < 11$
B) $12 < t < 15$
C) $0 < t < 4$
D) $4 < t < 7$

Quizplus · Accepted Answer

The equation for the height $h$ of the flare at time $t$ is given by $h = -16t^2 + 176t + 448$, where $v_0 = 176 \, \text{ft/s}$ is the initial velocity and $h_0 = 448 \, \text{ft}$ is the initial height (depth of the gorge). To find when the flare is above the rim (height $h > 0$), we set $h = 0$ and solve for $t$:$$0 = -16t^2 + 176t + 448.$$This is a quadratic equation in the form of $at^2 + bt + c = 0$, where $a = -16$, $b = 176$, and $c = 448$. Solving this equation using the quadratic formula gives the times when the flare is at the rim level. However, since the flare starts from below the rim and goes up, we are interested in the interval when it is above the rim, which is between the two roots of this equation.Given the specific choices, without explicitly solving the quadratic equation, we know the flare must rise above the rim after being fired and then fall back down, creating an interval during which it is visible. Choice D) $4 < t < 7$ correctly suggests a reasonable time interval for the flare to be visible above the rim, considering its initial velocity and the depth of the gorge. This choice implies that the flare takes some time to reach its peak height above the rim and then falls back down, fitting the expected behavior of a projectile.

Solve the Problem 176ft/sec176 \mathrm { ft } / \mathrm { sec }176ft/sec

Solve the Problem $176 \mathrm { ft } / \mathrm { sec }$