Jacob and David devise an experiment to determine how tall the school building is. They stand on the roof of the school where Jacob drops a rock to the ground while David times it. From the time Jacob drops the rock to the time they hear it hit the ground, it takes 0.81 s. If the speed of sound is 340 m/s how tall is the school? Remember that it takes time for the rock to fall to the ground AND time for the sound of it to come back up for them to hear it.

You need to equations to solve this question. You need to know the time it takes to reach the ground and the time it takes for the two observers to hear it. The distance in both cases is the same, so the two conditions can be equated to each other.

Equations.

d = the height of the buildings

a is the acceleration due to gravity and is a constant 9.8i1 m/s^2

t1 is the time it takes to hit the ground

The initial velocity = 0 because the rock was dropped

Equation One

d = vi*t + 1/2 a t^2

Equation Two

d = v * t2

v is the velocity of sound

t2 is the time it takes for the sound to come back from hitting the ground.

d is the height of the school

Solution

vi*t + 1/2 a t1 ^2 = v * t2

t2 = 0.81 - t1

The times are separate, but they add up to 0.81 so

t1 I+ t2 = 0.81

t2 = 0.81 - t1

Rewrite the equation putting in the givens. Remember vi = 0

1/2 9.81 t1^2 = 340 * (0.81 - t1) Remove the brackets

4.91* t1^2 = 275.4 - 340 t1 Transfer the right side to the left

4.91*t1^2 + 340t1 - 275.4 = 0 Use the quadratic formula to solve

a = 4.91

b = 240

c = -275.4

t1 = (-240 +/- sqrt(240^2 - 4*4.91*-275.4))/(2*4.91)

t1 = .8007 = 0.80

t1 + t2 = 0.81

0.80 + t2 = 0.81 - 0.80

t2 = 0.01

What you have found is the time it takes for the sound to come back to the two observers which is 0.01 seconds

d(height of school) = 340 * 0.01 = 3.4 meters