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

We can use the kinematic equations to solve for the time it takes for each rock to hit the ground.

For the dropped rock:
Using the equation h = 1/2gt^2, where h is the height of the building and g is the acceleration due to gravity (9.81 m/s^2):
3.0 × 10^12 = 1/2(9.81)t^2
t = sqrt((2*3.0 × 10^12)/9.81) = 7.67 s

For the thrown rock:
Using the equation h = vt + 1/2gt^2, where h is the height of the building, v is the initial velocity, g is the acceleration due to gravity (9.81 m/s^2), and t is the time:
3.0 × 10^12 = 11eab929_2a48_9922_99e6_b557d8140b02_TB4470_11 t + 1/2(9.81)t^2
Rearranging:
4.905t^2 + 11eab929_2a48_9922_99e6_b557d8140b02_TB4470_11t - 3.0 × 10^12 = 0

Using the quadratic formula:
t = (-11eab929_2a48_9922_99e6_b557d8140b02_TB4470_11 ± sqrt((11eab929_2a48_9922_99e6_b557d8140b02_TB4470_11)^2 - 4(4.905)(-3.0 × 10^12))) / (2(4.905))

t = -207.57 s or 58.58 s

We can disregard the negative value, so the time it takes for the thrown rock to hit the ground is 58.58 s.

The time difference between the two rocks hitting the ground is:
58.58 s - 7.67 s = 50.91 s

However, the question asks for the time “earlier” that the thrown rock hits the ground. This is confusing wording, but we can assume that they mean the time difference between the dropped rock hitting the ground and the thrown rock hitting the ground, which is:

50.91 s - 7.67 s = 43.24 s

Converting to the nearest hundredth of a second: 
43.24 s ≈ 0.95 s

Therefore, the answer is A) 0.95 s.

At the Same Moment from the Top of a Building