Physics, Free Fall Exercises. Part Two

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 t₁ the time it takes the stone to reach the ground on free fall and t₂ the time it takes the sound wave to travel from the ground to the top of the building, then t₁+t₂ = 6.5 s

The height h of the building is obtained from the general formula
y=y_i + v_it + ½ at², with
y=h, v_i=0, a=9,8 m/s².
Then h=½ gt₁²=½ ·(9.8 m/s²)t₁²=(4.9 m/s²)t₁²

Also, h = (340 m/s)·t₂. Equating the two equations and replacing t₂ by (6.5 - t₁) to calculate t₁:
(4.9 m/s²)t₁²= (340 m/s)·t₂ = (340 m/s)·(6.5 - t₁) or 4.9·t₁²= 2210 - 340t₁, with t₁ in seconds,

4.9·t₁² + 340t₁ - 2210 = 0

The solution of this quadratic equation is 5.98 s.

Then, the height of the building is h = (4.9 m/s²)(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/s² 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_it + ½ at² with the same vertical axis already mentioned, y_i = 0, v_i = +14 m/s, a = -9.8 m/s², we get

y_max = (14 m/s)·(1.43s) - (4.9m/s²)(1.43 s)² = 10 m

b) Keeping the same axis as before, and using v_i = +14 m/s, a = -9.8 m/s² and t = 4.5 s in y = y_i + v_it + ½ at², y = (14 m/s)·4.5 s - 4.9m/s²(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/s²·4.5 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