Question

A brick is dropped from the roof of a tall building. after it has been falling for a few seconds, it falls 47.0 m in a 1.00-s time interval. what distance will it fall during the next 1.00 s? ignore air resistance.

Answer 1

The brick falls with an average speed of 47m/s during 1 second interval

As the acceleration is constant, this will be our reference value at t = 0.5 s, so

lets calculate the value at the end of the interval

Vend_interval = Vinitial + g*(0.5s) = 47[m/s] +9.8(0.5)[m/s] =51.9[m/s]

With this speed we can calculate the distance traveled with the following expression

d = Vinitial*t + 1/2 * g*(t**2)

after 1 second t = 1s

d = 51.9 [m/s] + 4.9 [m/s] = 56.8 [m] (new distance traveled after one second)

Answer 2

Answer: 56.8 m

The object would fall with uniform acceleration (
`g=9.8 m/s^2`) as the brick was dropped.

In the given interval, velocity,
`u=(distance)/(time)=(47.0m)/(1.00s)=47.0m/s`

This would be the initial velocity for next 1 s. Insert the values in equation of motion:

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

where, s is the displacement, a is the acceleration and t is the time.

`s=47.0 m/s* 1.0 s+9.8 m/s^2* (1.0s)^2=56.8 m`