Question

a baseball dropped from the roof of a tall building takes 3.1 seconds to hit the ground. how tall is the building?

Answer 1

The height of the building is $\boxed{\bf 47.1\ \text{\bf meters}}$.

Further explanation:

Motion is defined as the change in position of an object over time.

It is described in terms of displacement, distance, velocity, acceleration, time and speed.

The relations between these quantities are known as the equations of motion.

The three equations of motion are as follows:

$\boxed{\begin{aligned}v&=u+at\\v^{2}&=u^{2}+2as\\s&=ut+\dfrac{1}{2}at^{2}\end{aligned}}$

Where, $s$ is the displacement, $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration and $t$ is time.

It is given that a baseball dropped from the roof of a tall building takes $3.1$ seconds that is $t=3.1\ \text{seconds}$.

Therefore, the initial velocity $u$ is $0$ and acceleration will be taken as acceleration due to gravity and that is $a=9.8\ m/s^{2}$.

Substitute the above values in equation $s=ut+\frac{1}{2}ay^{2}$ to obtain the height $s$ of the building.

$\begin{aligned}s&=(0\times3.1)+\left(\dfrac{1}{2}\times9.8\times(3.1)^{2}\right)\\&=0+\left(\dfrac{1}{2}\times9.8\times9.61\right)\\&=47.089\\ &\approx47.1\end{aligned}$  

Therefore, the height of the building is $\boxed{\bf 47.1\ \text{\bf meters}}$

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Answer details:

Grade: Middle school

Subject: Mathematics

Topic: Laws of motion

Keywords: Equation of motion, displacement, acceleration, distance, speed, time, initial velocity, final velocity, Newton laws of motion, circular motion.

Answer 2

Initial velocity = 0 m/s
Acceleration = 9.8 m/s²
Time = 3.1 s

s = ut + 1/2 at²
s = 1/2 x 9.8 x 3.1²
s = 47.1 m

The building is 47.1 meters tall