Solve the problem.
-An object is dropped from $9 \mathrm { ft }$ above the surface of the moon. How long will it take the object to hit the surface of the moon if $\mathrm { d } ^ { 2 } \mathrm {~s} / \mathrm { dt } ^ { 2 } = - 5.2 \mathrm { ft } / \mathrm { sec } ^ { 2 }$ ?
A) 3.46 sec
B) 1.32 sec
C) 0.87 sec
D) 1.86 sec

Question

Quizplus · Accepted Answer

The equation of motion under constant acceleration is $s = ut + \frac{1}{2}at^2$, where $s$ is the distance, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. Since the object is dropped, $u = 0$, and $s = 9$ ft, $a = -5.2$ ft/sec$^2$ (negative because it's directed downwards). Plugging these values into the equation gives $9 = 0 + \frac{1}{2}(-5.2)t^2$, solving for $t$ gives $t = 1.86$ sec.

Solve the Problem 9ft9 \mathrm { ft }9ft Above the Surface of the Moon

Solve the Problem $9 \mathrm { ft }$ Above the Surface of the Moon