Question

A brick is dropped from the roof of a tall building. After it has been falling for a few seconds, it falls 49.0 m in a 1.00-s time interval. What distance will it fall during the next 1.00s? ignore air resistance.

Answer 1

The problem requires using the kinematic equations for uniformly accelerated motion to find the total distance a brick will fall in the next second after already having fallen 49.0 m in the previous second.

Step-by-step explanation:

The student's question pertains to the motion of an object under the influence of gravity, specifically asking about the distance a brick will fall in the second after falling for a few seconds and already having fallen 49.0 m during one second.

To approach this problem, we can use the kinematic equations for uniformly accelerated motion, with the acceleration due to gravity (g) being approximately 9.8 m/s2. The equation s = ut + 0.5at2 can be used, where s is the distance, u is the initial velocity, a is the acceleration, and t is time.

Since the brick has already been falling, we must find its velocity after falling for the first few seconds before we can determine the distance it will fall in the next second.

Given that the brick falls 49.0 m in one second, we can infer that its initial velocity at the start of that second was such that, when combined with the acceleration due to gravity, it results in the observed distance fallen.

To find the initial velocity for the next second, we can use the final velocity equation v = u + at, and then apply the distance equation again to find the distance it will fall in the subsequent second.