**Free Fall Exercises**

**Exercise Two, Calculations Combining Free
Fall and Sound Wave Velocity.-** A stone is dropped from the top of a building.
The sound of the stone hitting the ground is heard 6.5 s later. If the sound
velocity is 340 m/s, calculate the height of the building.

**Solution.-** Calling t_{1} the
time it takes the stone to reach the ground on free fall and t_{2}
the time it takes the sound wave to travel from the ground to the top of the
building, then t_{1}+t_{2} = 6.5 s

The height h of the building is obtained from
the general formula

y=y_{i} + v_{i}t + ½ at^{2}, with

y=h, v_{i}=0, a=9,8 m/s^{2}.

Then h=½ gt_{1}^{2}=½
·(9.8 m/s^{2})t_{1}^{2}=(4.9 m/s^{2})t_{1}^{2}

Also, h = (340 m/s)·t_{2}. Equating
the two equations and replacing t_{2} by (6.5 - t_{1}) to
calculate t_{1}:

(4.9 m/s^{2})t_{1}^{2}= (340 m/s)·t_{2} =
(340 m/s)·(6.5 - t_{1}) or 4.9·t_{1}^{2}= 2210 - 340t_{1},
with t_{1} in seconds,

4.9·t_{1}^{2}
+ 340t_{1} - 2210 = 0

The solution of this quadratic equation is 5.98 s.

Then, the height of the building is h = (4.9
m/s^{2})(5.98s)^{2} = 175 m

**Exercise Three. Free Fall Combining Initial
Velocity and Time to Reach Ground.-** A man standing at the top of a building
throws a ball vertically upward with a velocity of 14 m/s. The ball reaches
the ground 4.5 s later. a) What is the maximum height reached by the ball?
b) How high is the building? c) With what velocity will it reach the ground?

**Solution.-**

a) Maximum height means v = 0. Then using v = v_{i} + at, and a vertical
axis with origin at the top of the building pointing upward, with

v = 0, a = -9.8 m/s^{2} and v_{i} = 14 m/s, the time to reach
maximum height is t = (14 m/s)/(9.8 s) = 1.43 s

Now, using y = y_{i} + v_{i}t
+ ½ at^{2} with the same vertical axis already mentioned, y_{i}
= 0, v_{i} = +14 m/s, a = -9.8 m/s^{2}, we get

y_{max} = (14 m/s)·(1.43s) - (4.9m/s^{2})(1.43
s)^{2} = 10 m

b) Keeping the same axis as before, and using
v_{i} = +14 m/s, a = -9.8 m/s^{2} and t = 4.5 s in y = y_{i}
+ v_{i}t + ½ at^{2}, y = (14 m/s)·4.5 s - 4.9m/s^{2}(4.5
s)^{2} = -36.2 m. The result is negative as expected since the ground
is in the negative side of the axis. The height of the building is then 36.2
m.

c) The velocity reaching the ground is v =
14 m/s - 9.8 m/s^{2}·4.5 s = -30.1 m/s.

**Other Free Fall Exercises:**

**Free Fall Exercises, Part One**

**Free Fall Exercises, Part Three**

**To See the Free Fall Physics Theory:**

**Go To Free Fall Theory**

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