At the same moment from the top of a building 3.0 × 10² m tall,one rock is dropped and one is thrown downward with an initial velocity of .Both of them experience negligible air resistance.How much EARLIER does the thrown rock strike the ground? A) 0.95 s B) 0.86 s C) 0.67 s D) They land at exactly the same time.

Question

Quizplus · Accepted Answer

First we need to find the time it takes for the dropped rock to hit the ground. We can use the equation:
$$d = \frac{1}{2}gt^2$$
where $d$ is the distance (height) of the building, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $t$ is the time. Plugging in the values, we get:
$$3.0 	imes 10^{12} = \frac{1}{2}(9.8)t^2$$
Solving for $t$, we get $t \approx 5.43$ seconds.

Now we need to find the time it takes for the thrown rock to hit the ground. We can use the equation:
$$d = vt + \frac{1}{2}gt^2$$
where $v$ is the initial velocity of the rock (downward in this case). Since the initial velocity is negative, we can use the magnitude and add a negative sign to the answer later. Plugging in the values, we get:
$$3.0 	imes 10^{12} = 11.83t - \frac{1}{2}(9.8)t^2$$
This is a quadratic equation, so we can use the quadratic formula to solve for $t$:
$$t = \frac{-11.83 \pm \sqrt{(11.83)^2 - 4(-\frac{1}{2}(9.8))(3.0 	imes 10^{12})}}{2(-\frac{1}{2}(9.8))}$$
Simplifying, we get:
$$t \approx 4.48 	ext{ seconds or } -0.41 	ext{ seconds}$$
Since we are looking for the time difference between the two rocks hitting the ground, we only need the positive root. Therefore, the thrown rock hits the ground 5.43 - 4.48 = 0.95 seconds earlier than the dropped rock. The answer is A.

At the Same Moment from the Top of a Building