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 \(v\), because no forces act in the horizontal direction. The speed in a uniform motion is given by
\(v=\frac{S}{t}\)
where S is the distance traveled and t the time taken.
In this case, the distance by the rock before hitting the ground is \(S=115 m\) and the time taken is \(t=3.7 s\), so the initial speed is given by
\(v=\frac{115 m}{3.7 s}=31.1 m/s\)
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 \(a=9.8 m/s^2\) (acceleration of gravity) towards the ground. In a uniformly accelerated motion, the distance traveled by the object is given by
\(S=\frac{1}{2}at^2\)
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:
\(S=\frac{1}{2}(9.8 m/s^2)(3.7 s)^2=67.1 m\)
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
\(h=\frac{67.1 m}{7}=9.6 m\)
And so, the height of the 28th floor is
\(h=28\cdot 9.6 m=268.8 m\)
We can find the total time of the fall in this case by using the same formula of the previous part:
\(S=\frac{1}{2}at^2\)
In this case, S=268.8 m, so we can re-arrange the formula to find t
\(t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(268.8 m)}{9.8 m/s^2}}=7.4 s\)
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
\(S=vt=(31.1 m/s)(7.4 s)=230.1 m\)
And this is how far from the building the rock lands.
