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A rock is thrown vertically upward with a speed of 12.0 m/s from the roof of a building that is 60.0 meters above the ground. Assume free fall.

How many seconds after being thrown does the rock strike the ground? What is the speed of the rock just before it strikes the ground?

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Final answer:

The time and final velocity of a rock thrown vertically upward from a building can be calculated using kinematic equations, considering the initial velocity, acceleration due to gravity, and the height of the building.

Step-by-step explanation:

Calculating Time and Speed of a Falling Rock

An important problem in physics is calculating the time it takes for an object, like a rock, to fall to the ground, and its final velocity. Here, we need to determine the time and velocity for a rock thrown vertically upward at 12.0 m/s from a 60.0 m building.

Step-by-step Calculation

First, we use the kinematic equation for vertical motion:
s = ut + ½ at². The initial velocity (u) is 12.0 m/s, the acceleration (a) due to gravity is -9.81 m/s² (negative because it's downward), and the distance (s) is -60.0 m (negative as the final position is below the starting point).

To calculate the time (t), we rearrange the equation:
0 = 12t - ½(9.81)t² - 60. Solving this quadratic equation gives two times: one for when the rock reaches its peak and a longer time for when it hits the ground.

For the final velocity (v), we use: v = u + at. Substituting the longer time calculated above and the known values, we find the speed just before the rock strikes the ground.

By applying these equations and solving, we get the time the rock takes to hit the ground and its final speed at that moment. Since air resistance is ignored in this calculation, the final speed will be the same whether the rock is thrown up or dropped directly from the same height.

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