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 10 m/s. 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

The time it takes for an object to fall a distance h from rest is given by the formula: 
t = sqrt(2h/g), where g is the acceleration due to gravity (9.8 m/s^2).

For the dropped rock: 
t = sqrt(2*3.0 × 10^12/9.8) = 244.9 seconds

For the thrown rock, the initial velocity is 10 m/s downwards, so its motion is given by:
h = vt + (1/2)at^2
where v is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time elapsed.

Solving for t:
3.0 × 10^12 = 10t + (1/2)(-9.8)t^2
7.84t^2 - 10t - 3.0 × 10^12 = 0
Using the quadratic formula, t = 344.9 seconds or -441.7 seconds. Since the time can't be negative, we discard that solution.

Thus, the thrown rock takes 344.9 seconds to hit the ground, while the dropped rock takes 244.9 seconds to hit the ground. The difference in time is:
344.9 - 244.9 = 100 seconds = 0.95 seconds (to two significant figures)

Therefore, the thrown rock hits the ground 0.95 seconds earlier than the dropped rock. Therefore, the answer is (A).

At the Same Moment from the Top of a Building