Solve the problem.
-On our moon, the acceleration of gravity is $1.6 \mathrm {~m} / \mathrm { sec } ^ { 2 }$. If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 45 seconds later?
A) $72 \mathrm {~m} / \mathrm { sec }$
B) $3240 \mathrm {~m} / \mathrm { sec }$
C) $- 36 \mathrm {~m} / \mathrm { sec }$
D) $- 72 \mathrm {~m} / \mathrm { sec }$

Question

Quizplus · Accepted Answer

Using the formula $d = \frac{1}{2} at^2$, where $d$ is the distance, $a$ is the acceleration, and $t$ is the time, we can solve for the distance the rock falls: $d = \frac{1}{2}(1.6 \mathrm{~m/sec^2})(45\mathrm{~sec})^2 = 1620 \mathrm{~m}$. Next, we can use the formula $v_f^2 = v_i^2 + 2ad$, where $v_f$ is the final velocity, $v_i$ is the initial velocity (which is 0), $a$ is the acceleration, and $d$ is the distance. Solving for $v_f$, we get $v_f = -\sqrt{2ad} = -\sqrt{2(1.6\mathrm{~m/sec^2})(1620\mathrm{~m})} = -71.6 \mathrm{~m/sec}$. Since the velocity is negative, it means the rock is traveling downwards, so the answer is D.

Solve the Problem 1.6 m/sec21.6 \mathrm {~m} / \mathrm { sec } ^ { 2 }1.6 m/sec2

Solve the Problem $1.6 \mathrm {~m} / \mathrm { sec } ^ { 2 }$