Suppose the position function for a free-falling object on a certain planet is given by $s ( t ) = - 14 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$. A silver coin is dropped from the top of a building that is 1370 feet tall. Find velocity of the coin at impact. Round your answer to the three decimal places.
A) $- 286.705 \mathrm { ft } / \mathrm { sec }$
B) $- 138.492 \mathrm { ft } / \mathrm { sec }$
C) $- 111.041 \mathrm { ft } / \mathrm { sec }$
D) $- 276.984 \mathrm { ft } / \mathrm { sec }$
E) $- 261.984 \mathrm { ft } / \mathrm { sec }$

Question

Quizplus · Accepted Answer

First, we need to find the initial velocity, $v_0$. Since the coin is dropped from rest, $v_0 = 0$. The acceleration due to gravity, $a$, is given by the coefficient of $t^2$ in the position function, which is $a = -14$ ft/s\textsuperscript{2}. The height from which the coin is dropped, $s_0$, is 1370 ft.

Now we can use the formula for velocity in terms of initial velocity, acceleration, and time:

$$v(t) = v_0 + at$$

We want to find the velocity at impact, which is the velocity when the coin hits the ground. We don't know how long it takes the coin to fall, but we can use the fact that at impact, $s(t)$ is equal to 0:

$$0 = -14t^2 + v_0t + s_0$$

Solving this quadratic equation for $t$, we get:

$$t = \frac{-v_0 \pm \sqrt{v_0^2 - 4(-14)(s_0)}}{2(-14)}$$

Since the coin is dropped from rest, we can simplify this to:

$$t = \frac{\sqrt{1370}}{\sqrt{14 \cdot 32}} \approx 1.303$$

Now we can plug this value of $t$ into the formula for velocity:

$$v_{\text{impact}} = 0 + (-14)(1.303) \approx \boxed{-276.984 \textrm{ ft/s}}$$

Therefore, the best choice is D, with the velocity at impact of approximately -276.984 ft/s.

Suppose the Position Function for a Free-Falling Object on a Certain