Question

A brick is dropped vertically from a building at 2 m/s to reach the ground after s Calculate the distance the brick fell.

Answer 1

To calculate the distance the brick fell, we can use the kinematic equation:

s = ut + (1/2)at^2

where:
s = distance
u = initial velocity (2 m/s, in this case)
t = time
a = acceleration (in this case, due to gravity, a = -9.8 m/s^2)

Since the brick is dropped vertically, the initial velocity is in the downward direction, so the acceleration due to gravity is negative.

Given that the brick reaches the ground after time "s", we can set t = s in the equation above.

s = us + (1/2)as^2
s = (2 m/s)(s) + (1/2)(-9.8 m/s^2)(s^2)
s = 2s - 4.9s^2

To solve for s, we can set the equation equal to 0 and use the quadratic formula:

4.9s^2 - 2s = 0

Factoring out s:
s(4.9s - 2) = 0

Setting each factor equal to 0:
s = 0 or 4.9s - 2 = 0

Ignoring the solution s = 0 (since the brick actually fell), we can solve the second equation:

4.9s - 2 = 0
4.9s = 2
s = 2/4.9

Therefore, the distance the brick fell is approximately 0.408 meters.