You throw a rock out of a 7th story window. you time that it takes 3.7 seconds to hit the ground, and measure that it hit the ground 115 m from the base of the building?
how fast must you have thrown the rock?
how high up must the 7th floor be?
if you had thrown it at the same speed but at the 28th floor, how far from the base of the building would it land?

1) 31.1 m/s

The rock has been thrown straight out of the window: its motion on the horizontal direction is simply a uniform motion, with constant speed , because no forces act in the horizontal direction. The speed in a uniform motion is given by

where S is the distance traveled and t the time taken.

In this case, the distance by the rock before hitting the ground is and the time taken is , so the initial speed is given by

2) 67.1 m

In this part of the problem we are only interested in the vertical motion of the rock. The vertical motion is a uniformly accelerated motion, with constant acceleration (acceleration of gravity) towards the ground. In a uniformly accelerated motion, the distance traveled by the object is given by

where t is the time. Substituting a=9.8 m/s^2 and t=3.7 s, we can find S, the vertical distance covered by the rock, which corresponds to the height of the 7th floor:

3) 230.1 m

The height of the 7th floor is 67.1 m. So we can assume that the height of each floor is

And so, the height of the 28th floor is

We can find the total time of the fall in this case by using the same formula of the previous part:

In this case, S=268.8 m, so we can re-arrange the formula to find t

And now we can consider the motion of the rock on the horizontal direction: we know that the rock travels at a constant speed of v=31.1 m/s, so the distance traveled is

And this is how far from the building the rock lands.

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